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Volume 9, Issue 1

# A Picone identity for variable exponent operators and applications

Rakesh Arora
/ Jacques Giacomoni
/ Guillaume Warnault
Published Online: 2019-05-16 | DOI: https://doi.org/10.1515/anona-2020-0003

## Abstract

In this work, we establish a new Picone identity for anisotropic quasilinear operators, such as the p(x)-Laplacian defined as div(|∇ u|p(x)−2u). Our extension provides a new version of the Diaz-Saa inequality and new uniqueness results to some quasilinear elliptic equations with variable exponents. This new Picone identity can be also used to prove some accretivity property to a class of fast diffusion equations involving variable exponents. Using this, we prove for this class of parabolic equations a new weak comparison principle.

MSC 2010: 35A16; 35B51; 35K92; 35J92

## 1 Introduction and main results

The main aim of this paper is to prove a new version of the Picone identity involving quasilinear elliptic operators with variable exponent. The Picone identity is already known for homogeneous quasilinear elliptic as p-Laplacian with 1 < p < ∞.

In [1], M. Picone considers the homogeneous second order linear differential system

$(a1(x)u′)′+a2(x)u=0 (b1(x)v′)′+b2(x)v=0$

and proved for differentiable functions u , v ≠ 0 the pointwise relation:

$(uv(a1u′v−b1uv′))′=(b2−a2)u2+(a1−b1)u′2+b1(u′−v′uv)2$(1.1)

and in [2], extended (1.1) to the Laplace operator, i.e. for differentiable functions u ≥ 0 , v > 0 one has

$|∇u|2+u2v2|∇v|2−2uv∇u.∇v=|∇u|2−∇(u2v).∇v≥0.$(1.2)

In [3], Allegretto and Huang extended (1.2) to the p-Laplacian operator with 1 < p < ∞. Precisely, for differentiable functions v > 0 and u ≥ 0 we have

$|∇u|p−|∇v|p−2∇v.∇(upvp−1)≥0.$

Picone identity plays an important role for proving qualitative properties of differential operators. In this regard, various attempts have been made to generalize Picone identity for different types of differential equations. At the same time, the study of differential equations and variational problems with variable exponents are getting more and more attention. Indeed, the mathematical problems related to nonstandard p(x)-growth conditions are connected to many different areas as the nonlinear elasticity theory and non-Newtonian fluids models (see [4, 5]). In particular the importance of investigating these kinds of problems lies in modelling various anisotropic features that occur in electrorheological fluids models, image restoration [6], filtration process in complex media, stratigraphy problems [7] and heterogeneous biological interactions [8]. The mathematical framework to deal with these problems are the generalized Orlicz Space Lp(x)(Ω) and the generalized Orlicz-Sobolev Space W1,p(x)(Ω). We refer to [9, 10, 11, 12, 13, 14] for the existence and regularity of minimizers in variational problems.

In [15, 16], several applications of Picone-type identity for p(⋅) = constant case have been obtained. This original identity is not further applicable for differential equations with p(x)-growth conditions. So, it is relevant to establish a new version of the Picone identity to include a large class of nonstandard p(x)-growth problems. In [5, 12, 17] convexity arguments to homogeneous functionals have been used to deal with quasilinear elliptic and parabolic problems with variable exponents. In the present paper, taking advantage of our new Picone pointwise identity, we give further applications in the context of elliptic and parabolic problems.

Before giving the statement of our main results, we first introduce notations and function spaces. Let Ω ⊂ ℝN, N ≥ 1. We recall some definitions of variable exponent Lebesgue and Sobolev spaces. Let 𝓟(Ω) be the set of all measurable function p : Ω → [1, ∞[ in N-dimensional Lebesgue measure. Define

$ρp(u) =def∫Ω|u|p(x) dx.Lp(x)(Ω)={u:Ω→R| u is measurable and ρp(u)<∞}$

endowed with the norm

$∥u∥Lp(x)=inf{σ>0:ρp(uσ)≤1}.$

The corresponding Sobolev space is defined as follows:

$W1,p(x)(Ω)={u∈Lp(x)(Ω):|∇u|∈Lp(x)(Ω)}$

endowed with the norm

$∥u∥W1,p(x)=∥u∥Lp(x)+∥∇u∥Lp(x)$

and $\begin{array}{c}{W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)={W}_{0}^{1,1}\left(\mathit{\Omega }\right)\cap {W}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\end{array}$

In the sequel, we assume that Ω satisfies:

• (Ω)

For N = 1, Ω is a bounded open interval and for N ≥ 2, Ω is a bounded domain whose the boundary ∂ Ω is a compact manifold of class C1,γ for some γ ∈ (0, 1) and satisfies the interior sphere condition at every point of ∂ Ω.

Throughout the paper, we also assume that pC1(Ω). In addition, we suppose that

$1

Then, $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)={\overline{{C}_{0}^{\mathrm{\infty }}\left(\mathit{\Omega }\right)}}^{{W}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)}\end{array}$. We also recall some well-known properties on Lp(x) spaces (see [14]).

#### Proposition 1.1

Let pL(Ω). Then for any uLp(x)(Ω) we have:

1. ρp(u/∥uLp(x)) = 1.

2. uLp(x) → 0 if and only if ρp(u) → 0.

3. Lpc(x)(Ω) is the dual space of Lp(x)(Ω) where we denote by pc the conjugate exponent of p defined as

$pc(x)=p(x)p(x)−1.$

Proposition 1.1 (i) implies that: if ∥uLp(x) ≥ 1,

$∥u∥Lp(x)p−≤ρp(u)≤∥u∥Lp(x)p+$(1.3)

and if ∥uLp(x) ≤ 1

$∥u∥Lp(x)p+≤ρp(u)≤∥u∥Lp(x)p−.$(1.4)

Moreover, we have also the generalized Hölder inequality: for p measurable function in Ω, there exists a constant C = C(p+, p) ≥ 1 such that for any fLp(x)(Ω) and gLpc(x)(Ω)

$∫Ω|f(x)g(x)| dx≤C∥f∥Lp(x)∥g∥Lpc(x).$(1.5)

In Section 2, we prove the Picone identity for a general class of nonlinear operator. More precisely, we consider a continuous operator A : Ω × ℝN → ℝ such that (x, ξ) → A(x, ξ) is differentiable with respect to variable ξ and satisfies:

• (A1)

ξA(x, ξ) is positively p(x)-homogeneous i.e. A(x, t ξ) = tp(x) A(x, ξ), ∀ t ∈ ℝ+, ξ ∈ ℝN and a.e. xΩ.

• (A2)]

ξA(x, ξ) is strictly convex for any xΩ.

#### Remark 1.1

From the assumptions of A, we deduce A(x, ξ) > 0 for ξ ≠ 0 and for any xΩ.

By using the convexity and the p(x)-homogeneity of the operator A, we prove the following extension of the Picone identity:

#### Theorem 1.1

(Picone identity). Let A : Ω × ℝN → ℝ is a continuous and differentiable function satisfying (A1) and (A2). Let v0, vL(Ω) belonging to $\begin{array}{}{\stackrel{˙}{V}}_{+}^{r}\text{\hspace{0.17em}}\stackrel{\text{def}}{=}\left\{v:\mathit{\Omega }\to \left(0,+\mathrm{\infty }\right)\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}{v}^{\frac{1}{r}}\in {W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\right\}\end{array}$ for some r ≥ 1. Then

$1p(x)〈∂ξA(x,∇v01/r),∇(vv0(r−1)/r)〉≤Arp(x)(x,∇v1/r)A(p(x)−r)p(x)(x,∇v01/r)$

where 〈., .〉 is the inner scalar product and the above inequality is strict if r > 1 or $\begin{array}{}\frac{v}{{v}_{0}}\end{array}$ ≢ Const > 0.

From the above Picone identity, we can show an extension of the famous Diaz-Saa inequality to the class of operators with variable exponent. This inequality is strongly linked to the strict convexity of some associated homogeneous energy type functional.

#### Theorem 1.2

(Diaz-Saa inequality). Let A : Ω × ℝN → ℝ is a continuous and differentiable function satisfying (A1) and (A2) and define $\begin{array}{}a\left(x,\xi \right)=\left({a}_{i}\left(x,\xi \right){\right)}_{i}\text{\hspace{0.17em}}\stackrel{\text{def}}{=}{\left(\frac{1}{p\left(x\right)}{\mathrm{\partial }}_{{\xi }_{i}}A\left(x,\xi \right)\right)}_{i}\end{array}$. Assume in addition that there exists Λ > 0 such that

$a∈C1(Ω×(RN∖{0}))N and ∑i,j=1N∂ai(x,ξ)∂ξj≤Λ|ξ|p(x)−2$

for all (x, ξ) ∈ Ω × ℝN ∖ {0}. Then, we have in the sense of distributions, for any r ∈ [1, p]

$∫Ω(−div(a(x,∇w1))w1r−1+div(a(x,∇w2))w2r−1)(w1r−w2r) dx≥0$(1.6)

for any w1, w2$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\end{array}$, positive in Ω such that $\begin{array}{}\frac{{w}_{1}}{{w}_{2}},\frac{{w}_{2}}{{w}_{1}}\in {L}^{\mathrm{\infty }}\left(\mathit{\Omega }\right)\end{array}$. Moreover, if the equality occurs in (1.6), then w1/w2 is constant in Ω. If p(x) ≢ r in Ω then even w1 = w2 holds in Ω.

In sections 3, 4 and 5, we derive some applications of the new Picone identity. Precisely, we investigate the solvability of some boundary problems involving quasilinear elliptic operators with variable exponent.

In section 3, we consider the following nonlinear problem:

$−Δp(x)u+g(x,u)=f(x,u) in Ω;u>0 in Ω;u=0 on ∂Ω.$(1.7)

The extended Picone identity can be reformulated as in Lemma 3.1 below. Together with the strong maximum principle and elliptic regularity, this identity can be used to prove the uniqueness of weak solutions to elliptic equations as (1.7). In particular, we establish the following result:

#### Theorem 1.3

Let f, g : Ω × [0, ∞) → ℝ+ be defined as f(x, t) = h(x)tq(x)−1 and g(x, t) = l(x)ts(x)−1 with 1 ≤ q, sC(Ω) such that

• q+ < p < s and q ≥ 1;

• h, lL(Ω), positive functions such that $\begin{array}{}x\to \frac{h\left(x\right)}{l\left(x\right)}\in {L}^{\mathrm{\infty }}\left(\mathit{\Omega }\right)\end{array}$.

Then, there exists a weak solution u to (1.7), i.e. u belongs to $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω) and satisfies for any ϕ$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω):

$∫Ω|∇u|p(x)−2∇u.∇ϕ dx=∫Ω(f(x,u)−g(x,u))ϕ dx.$

Furthermore uC1,α(Ω) for some α ∈ (0, 1) and $\begin{array}{}0\le {u}^{{s}_{-}-{q}_{+}}\le max\left\{\parallel \frac{h}{l}{\parallel }_{{L}^{\mathrm{\infty }}},1\right\}\end{array}$ a.e. in Ω.

Assume in addition that $\begin{array}{}x\to \frac{l\left(x\right)}{h\left(x\right)}\end{array}$ belongs to L(Ω), then $\begin{array}{}u\in {C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\text{\hspace{0.17em}}\stackrel{\text{def}}{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{v\in {C}_{0}\left(\overline{\mathit{\Omega }}\right)\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}\mathrm{\exists }\phantom{\rule{thinmathspace}{0ex}}{c}_{1},\phantom{\rule{thinmathspace}{0ex}}{c}_{2}\in {\mathbb{R}}_{\ast }^{+}:\phantom{\rule{thinmathspace}{0ex}}{c}_{1}\le \frac{v}{\text{dist}\left(x,\mathrm{\partial }\mathit{\Omega }\right)}\le {c}_{2}\right\}\end{array}$ and is the unique weak solution to (1.7).

Regarding the current literature, Theorem 1.3 does not require any subcritical growth condition for g to establish existence and uniqueness of the weak solution to (1.7).

In section 4, we study a nonlinear fast diffusion equation (F.D.E. for short) driven by p(x)-Laplacian. From the physical Fick’s law, the diffusion coefficient of our problem is then proportional to |∇ u(x, t)|p(x)−2. It naturally leads to investigate the following F.D.E. type problem:

$q2q−1∂t(u2q−1)−Δp(x)u=f(x,u)+h(t,x)uq−1 in QT; u>0 in QT; u=0 on Γ; u(0,.)=u0 in Ω$(1.8)

where q ∈ (1, p), QT = (0, T) × Ω and Γ = (0, T) × ∂ Ω for some T > 0. We suppose that hL(QT) and nonnegative. The assumptions on f are given by

• (f1)

f : Ω × ℝ+ → ℝ+ is a function such that f(x, 0) = 0 for all xΩ and f ≢ 0;

• (f2)

$\begin{array}{}\underset{s\to {0}^{+}}{lim}\frac{f\left(x,s\right)}{{s}^{2q-1}}=+\mathrm{\infty }\end{array}$ uniformly in x;

• (f3)

for any xΩ, s$\begin{array}{}\frac{f\left(x,s\right)}{{s}^{q-1}}\end{array}$ is nonincreasing in ℝ+ ∖ {0}.

#### Remark 1.2

Conditions (f1) and (f3) imply there exist positive constant C1, C2 such that for any (x, s) ∈ Ω × ℝ+:

$0≤f(x,s)≤C1+C2sq−1,$

i.e. f has a strict subhomogeneous growth.

We set 𝓡 the operator defined by $\begin{array}{}\mathcal{R}v=\frac{-{\mathit{\Delta }}_{p\left(x\right)}\left({v}^{1/q}\right)}{{v}^{\left(q-1\right)/q}}-\frac{f\left(x,{v}^{1/q}\right)}{{v}^{\left(q-1\right)/q}}\end{array}$ and the associated domain

$D(R)={v:Ω→(0,∞):v1/q∈W01,p(x)(Ω),v∈L2(Ω),Rv∈L2(Ω)}.$

Note that 𝓓(𝓡) contains for instance solutions to (4.18). One can also easily check that solutions to (4.19) belong to 𝓓(𝓡)L2(Ω). In the sequel, we denote $\begin{array}{}{X}^{+}\text{\hspace{0.17em}}\stackrel{\text{def}}{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{x\in X\phantom{\rule{thinmathspace}{0ex}}|\phantom{\rule{thinmathspace}{0ex}}x\ge 0\right\}\end{array}$ the associated positive cone of a given real vector space X.

In order to establish existence and properties of weak solutions to (1.8), we investigate the following related parabolic problem:

$vq−1∂t(vq)−Δp(x)v=h(t,x)vq−1+f(x,v) in QT; v>0 in QT; v=0 on Γ; v(0,.)=v0(x)>0 in Ω.$(1.9)

The notion of weak solution for (1.9) is given as follows:

#### Definition 1.1

A weak solution to (1.9) is any positive function

vL(0, T; $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω)) ∩ L(QT) ∩ C(0, T; Lr(Ω)) for any r ≥ 1 such that t(vq) ∈ L2(QT) and for any ϕ$\begin{array}{}{C}_{0}^{\mathrm{\infty }}\end{array}$(QT) satisfies

$∫0T∫Ω∂t(vq)vq−1ϕ dxdt+∫0T∫Ω|∇v|p(x)−2∇v.∇ϕ dxdt=∫0T∫Ωh(t,x)vq−1ϕ dxdt+∫0T∫Ωf(x,v)ϕ dxdt.$(1.10)

Concerning (1.9), we prove the following results:

#### Theorem 1.4

Let T > 0, v0$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\cap {W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\end{array}$. In addition, there exists h0L(Ω), h0 ≢ 0 and h(t, x) ≥ h0(x) ≥ 0 for a.e xΩ, for a.e. t ≥ 0. Assume in addition q$\begin{array}{}min\left\{\frac{N}{2}+1,{p}_{-}\right\}\end{array}$ and f satisfies (f1) − (f3). Then there exists a weak solution to (1.9).

Based on the accretivity of 𝓡 with domain 𝓓(𝓡), we show the following result providing a contraction property for weak solutions to (1.9) under suitable conditions on initial data:

#### Theorem 1.5

Let v1 and v2 are weak solutions of (1.9) with initial data u0, v0$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\cap {W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\end{array}$ and such that $\begin{array}{}{u}_{0}^{q},\text{\hspace{0.17em}}{v}_{0}^{q}\in {\overline{\mathcal{D}\left(\mathcal{R}\right)}}^{{L}^{2}\left(\mathit{\Omega }\right)}\end{array}$ and h, gL(QT), such that hh0, gg0 with h0, g0 as in Theorem 1.4. Then, for any 0 ≤ tT,

$∥(v1q(t)−v2q(t))+∥L2≤∥(u0q−v0q)+∥L2+∫0t∥(h(s)−g(s))+∥L2ds.$(1.11)

Furthermore, using a similar approach as in [8], we consider for ϵ > 0 the perturbed operator 𝓡ϵ v = $\begin{array}{}\frac{-{\mathit{\Delta }}_{p\left(x\right)}\left({v}^{1/q}\right)}{\left(v+ϵ{\right)}^{\left(q-1\right)/q}}-\frac{f\left(x,{v}^{1/q}\right)}{\left(v+ϵ{\right)}^{\left(q-1\right)/q}}\end{array}$. If p ≥ 2, we can prove (as in Proposition 2.6 in [8]) that

$D(Rϵ)¯L2(Ω)⊃V˙+q∩Cdq0(Ω¯)+.$

Arguing as in Theorem 1.5 with the operator 𝓡ϵ instead of 𝓡 and passing to the limit as ϵ → 0+, we get:

#### Corollary 1.1

Assume p ≥ 2. Let v1 and v2 are weak solutions of (1.9) with initial data u0, v0$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\cap {W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\end{array}$. Then Theorem 1.5 holds.

From Theorem 1.5, we derive the following comparison principle from which uniqueness of the weak solution to problem (1.9) follows:

#### Corollary 1.2

Let u and v are the weak solutions of (1.9) with initial data u0, v0 satisfying conditions in Theorem 1.5 or Corollary 1.1. Assume u0v0 and h, gL(QT) , h0L(Ω) such that and 0 < h0hg. Then uv.

#### Remark 1.3

If vL(QT)+ then from Proposition 9.5 in [18] we obtain $\begin{array}{}\frac{q}{2q-1}\end{array}$ t(v2q−1) = vq−1 t(vq) = q v2q−2 t v in weak sense.

#### Remark 1.4

From Theorem 1.5, we can derive stabilization results for the evolution equation (1.8) in Lq(Ω) with q ∈ [2, ∞) (see [12] in this regard).

From the above remark, under assumptions given in Theorem 1.4, we obtain the existence of weak solutions to (1.8) satisfying the monotonicity properties in Theorem 1.5 and Corollaries 1.1, 1.2. We highlight that in our knowledge there is no result available in the current literature about F.D.E. with variable exponent. In this regard our results are completely new.

In the previous applications, the condition (A1) plays a crucial role to get suitable convexity property of energy functionals. In section 5, we study a quasilinear elliptic problem where this condition is not satisfied. Precisely, given ϵ > 0, we study the following nonhomogeneous quasilinear elliptic problem:

$−div((|∇u|2+ϵu2)p(x)−22∇u)−(|∇u|2+ϵu2)p(x)−22ϵu=g(x,u) in Ω; u=0 on ∂Ω; u>0 in Ω$(1.12)

where g satisfies (f1) and () for some m ∈ [1, p]:

• (g̃)

For any xΩ, s$\begin{array}{}\frac{g\left(x,s\right)}{{s}^{m-1}}\end{array}$ is decreasing in ℝ+ ∖ {0} and a.e. in Ω.

Then we prove the following result:

#### Theorem 1.6

Assume that g satisfies (f1) and (). Then for any ϵ, (1.12) admits one and only one positive weak solution. Furthermore, uC1(Ω), u > 0 in Ω and $\begin{array}{}\frac{\mathrm{\partial }u}{\mathrm{\partial }\stackrel{\to }{n}}\end{array}$ < 0 on ∂ Ω.

To get the uniqueness result contained in Theorem 1.6, we exploit the hidden convexity property of the associated energy functional in the interior of positive cone of C1(Ω).

## 2.1 Picone identity

First we recall the notion of strict ray-convexity.

#### Definition 2.1

Let X be a real vector space. Let be a non empty cone in X. A function J : → ℝ is ray-strictly convex if for all v1, v2 and for all θ ∈ (0, 1)

$J((1−θ)v1+θv2)≤(1−θ)J(v1)+θJ(v2)$

where the inequality is always strict unless v1 = Cv2 for some C > 0.

Then we have the following result:

#### Proposition 2.1

Let A satisfying (A1) and (A2) and let r ≥ 1. Then, for any xΩ the map ξNr(x, ξ) $\begin{array}{}\stackrel{\text{def}}{=}\end{array}$ A(x, ξ)r/p(x) is positively r-homogeneous and ray-strictly convex. For r > 1, ξNr(x, ξ) is even strictly convex.

#### Proof

We begin by the case r = 1. For any t ∈ ℝ+, we have N1(x, t ξ) = t N1(x, ξ). Furthermore,

$A(x,(1−t)ξ1+tξ2)≤(1−t)A(x,ξ1)+tA(x,ξ2)≤max{A(x,ξ1),A(x,ξ2)}$

for any xΩ, ξ1, ξ2 ∈ ℝN and t ∈ [0, 1]. Therefore

$N1(x,(1−t)ξ1+tξ2)≤max{N1(x,ξ1),N1(x,ξ2)}$(2.1)

and this inequality is always strict unless ξ1 = λ ξ2, for some λ > 0.

Now we prove that N1 is subadditive.

Without loss of generality, we can assume that ξ1 ≠ 0 and ξ2 ≠ 0 . Then we have N1(x, ξ1) > 0 and N1(x, ξ2) > 0. Therefore, from (2.1) and 1-homogeneity of N1(x, ξ) we obtain for any t ∈ (0, 1):

$N1(x,(1−t)ξ1N1(x,ξ1)+tξ2N1(x,ξ2))≤1.$

We now fix t such that

$1−tN1(x,ξ1)=tN1(x,ξ2) i.e. t=N1(x,ξ2)N1(x,ξ1)+N1(x,ξ2)≤1.$

Then we get

$N1(x,ξ1+ξ2N1(x,ξ1)+N1(x,ξ2))≤1$

and by 1-homogeneity of N1, we obtain

$N1(x,ξ1+ξ2)≤N1(x,ξ1)+N1(x,ξ2),i.e.N1 is subadditive.$

Finally for t ∈ (0, 1), ξ1λ ξ2, ∀ λ > 0

$N1(x,(1−t)ξ1+tξ2)

This proves that ξN1(x, ξ) is ray-strictly convex. Now consider the case r > 1. Since for any xΩ, ξ$\begin{array}{}{N}_{r}^{1/r}\end{array}$(x, ξ) = N1(x, ξ) is ray-strictly convex and thanks to the strict convexity of ttr on ℝ+, we deduce that ξNr(x, ξ) = $\begin{array}{}{N}_{1}^{r}\end{array}$(x, ξ) is strictly convex when r > 1. □

From Proposition 2.1 and from the r-homogeneity of Nr, we easily deduce the following convexity property of the energy functional:

#### Proposition 2.2

Under hypothesis of Proposition 2.1 and assume in addition A is continuous on Ω × ℝN. Then, for 1 ≤ r < p:

$V˙+r∩L∞(Ω)∋v→∫ΩA(x,∇(v1/r))dx$

is ray-strictly convex (if r > 1, it is even strictly convex).

#### Proof

We know that ξNr(x, ξ) = Ar/p(x)(x, ξ) is r-positively homogeneous and strictly convex if r > 1 and for r = 1 this function is ray-strictly convex. For v1, v2$\begin{array}{}{\stackrel{˙}{V}}_{+}^{r}\end{array}$ and θ ∈ (0, 1) define v = (1 − θ) v1 + θ v2 and we get

$Nr(x,∇vv)≤(1−θ)v1vNr(x,∇v1v1)+θv2vNr(x,∇v2v2).$

By homogeneity,

$Nr(x,∇(v1/r))≤(1−θ)Nr(x,∇(v11/r))+θNr(x,∇(v21/r))$

and equality holds if and only if v1 = λ v2 for some λ > 0. Using the convexity of ttp(x)/r for 1 ≤ r < p we obtain

$∫ΩA(x,∇v1/r) dx≤(1−θ)∫ΩA(x,∇v11/r) dx+θ∫ΩA(x,∇v21/r) dx.$

Moreover, if p(x) ≠ r equality holds if and only if v1 = v2. □

From Proposition 2.1, we deduce the proof of Picone identity.

#### Proof of Theorem 1.1

Firstly, we deal with the case r > 1. Then from Proposition 2.1, for any xΩ the function ξNr(x, ξ) = A(x, ξ)r/p(x) is strictly convex. Let ξ , ξ0 ∈ ℝN ∖ {0} such that ξξ0 then

$Nr(x,ξ)−Nr(x,ξ0)>〈∂ξNr(x,ξ0),ξ−ξ0〉.$

Setting $\begin{array}{}\stackrel{~}{a}\left(x,\xi \right)=\frac{1}{r}{\mathrm{\partial }}_{\xi }{N}_{r}\left(x,\xi \right)\end{array}$, we obtain:

$Nr(x,ξ)−〈a~(x,ξ0),ξ0〉>r〈a~(x,ξ0),ξ−ξ0〉.$

Let v, v0 > 0 and replacing ξ, ξ0 by ξ/v and ξ0/ v0 respectively in the above expression, we get

$Nr(x,ξv)>r〈a~(x,ξ0v0),ξv−r−1rξ0v0〉.$

Taking ξ = ∇ v and ξ0 = ∇ v0 and using (r − 1)-homogeneity of ã(x, .),

$N(x,∇vrv(r−1)/r)>1v0(r−1)/r〈a~(x,∇v0rv0(r−1)/r),∇v−r−1r∇v0v0v〉$

where the inequality is strict unless $\begin{array}{}\frac{\mathrm{\nabla }v}{v}=\frac{\mathrm{\nabla }{v}_{0}}{{v}_{0}}\end{array}$.

Since v1/r, $\begin{array}{}{v}_{0}^{1/r}\in {W}^{1,p\left(x\right)}\end{array}$(Ω) ∩ L(Ω), we can write

$∇(v1/r)=∇vrv(r−1)/r and ∇(vv0(r−1)/r)=1v0(r−1)/r(∇v−r−1r∇v0v0v)$

and we obtain

$N(x,∇v1/r)>〈a~(x,∇v01/r),∇(vv0(r−1)/r)〉.$(2.2)

We have

$a~(x,∇v01/r)=1r∂ξN(x,∇v01/r)=1r∂ξAr/p(x)(x,∇v01/r)=1p(x)∂ξA(x,∇v01/r)Ar−p(x)p(x)(x,∇v01/r)$

and by replacing in (2.2) we obtain

$Arp(x)(x,∇v1/r)Ap(x)−rp(x)(x,∇v01/r)>1p(x)〈∂ξA(x,∇v01/r),∇(vv0(r−1)/r)〉.$

Now we deal with the case r = 1. Let ξ , ξ0 ∈ ℝN ∖ {0} such that for any λ > 0, ξλ ξ0. Then, from Proposition 2.1, we have that

$N(x,ξ)−N(x,ξ0)≥〈∂ξN(x,ξ0),ξ−ξ0〉.$

Taking ξ = ∇ v and ξ0 = ∇ v0, we deduce

$N(x,∇v)−N(x,∇v0)≥〈∂ξN(x,∇v0),∇(v−v0)〉$

and

$A1p(x)(x,∇v)Ap(x)−1p(x)(x,∇v0)≥1p(x)〈∂ξA(x,∇v0),∇v〉$

for any xΩ and the inequality is strict unless v = λ v0 for some λ > 0. □

The Picone identity also holds for anisotropic operators of the following type:

$∑i=1N∇i(bi(x,∇iu))=∑i=1N∂∂xibix,∂u∂xi.$

Precisely we have:

#### Corollary 2.1

Let B : Ω × ℝ → ℝN is a continuous and differentiable function such that B(x, s) = (Bi(x, s))i=1,2,…N satisfying for any i, for any xΩ, the map sBi(x, s) is pi(x)-homogeneous and strictly convex with 1 < $\begin{array}{}{p}_{i}^{-}\le {p}_{i}\left(\cdot \right)\le {p}_{i}^{+}\end{array}$ < ∞. For any i, we define bi(x, s) = $\begin{array}{}\frac{1}{{p}_{i}\left(x\right)}\end{array}$ s Bi (x, s). Then, for v, v0$\begin{array}{}{\stackrel{˙}{V}}_{+}^{r}\end{array}$L(Ω), we have

$∑i=1Nbi(x,∂xi(v01/r))∂xivv0r−1r≤∑i=1NBirpi(x)x,∂xi(v1/r)Bipi(x)−rpi(x)x,∂xi(v01/r).$

#### Proof

By taking A(x, s) = Bi(x, s) in Theorem 1.1, we obtain ∀ i ∈ {1, 2, …, N}

$1pi(x)∂sBi(x,∂xi(v01/r)).∂xivv0r−1r≤Birpi(x)(x,∂xi(v1/r)).Bipi(x)−rpi(x)(x,∂xi(v01/r))$

for all v, v0$\begin{array}{}{\stackrel{˙}{V}}_{+}^{r}\end{array}$L(Ω) and i = 1, 2, …, N.

Then by summing the expression over i = 1, 2, …, N, we obtain

$∑i=1Nbi(x,∂xi(v01/r)).∂xivv0r−1r≤∑i=1NBirpi(x)(x,∂xi(v1/r)).Bipi(x)−rpi(x)(x,∂xi(v01/r)).$

## 2.2 An extension of the Diaz-Saa inequality

We prove the first application of Picone identity.

#### Proof of Theorem 1.2

The Picone identity implies

$Ar/p(x)(x,∇w1)A(p(x)−r)/p(x)(x,∇w2)≥a(x,∇w2).∇(w1rw2r−1).$

Using the Young inequality for r ∈ [1, p], we get

$rp(x)(A(x,∇w1)−A(x,∇w2))+A(x,∇w2)≥a(x,∇w2).∇(w1rw2r−1).$

Noting that for any ξ ∈ ℝN, A(x, ξ) = a(x, ξ). ξ, we deduce

$a(x,∇w2).∇(w2−w1rw2r−1) dx≥rp(x)(A(x,∇w2)−A(x,∇w1)).$(2.3)

Commuting w1 and w2, we have

$a(x,∇w1).∇(w1−w2rw1r−1)≥rp(x)(A(x,∇w1)−A(x,∇w2)).$(2.4)

Summing (2.3) and (2.4) and integrating over Ω yield

$∫Ωa(x,∇w1).∇(w1r−w2rw1r−1) dx+∫Ωa(x,∇w2).∇(w2r−w1rw2r−1)≥0.$

The rest of the proof is the consequence of Proposition 2.2. □

Diaz-Saa inequality also holds for anisotropic operators. Here we require that ξBi(x, ξ) is pi(x)-homogeneous and strictly convex and bi(x, ξ) = $\begin{array}{}\frac{1}{{p}_{i}\left(x\right)}\end{array}$ i Bi (x, ξ) where r ∈ ℝ, 1 ≤ r ≤ mini=1,2,…,N {(pi)}.

#### Corollary 2.2

Under the assumptions of Corollary 2.1 and in addition that there exist Λ > 0 such that for each i, $\begin{array}{}\left|\frac{\mathrm{\partial }{b}_{i}}{\mathrm{\partial }s}\left(x,s\right)\right|\le \mathit{\Lambda }|s{|}^{p\left(x\right)-2}\end{array}$. Then we have in the sense of distributions, for r ∈ [1, mini {(pi)}] and v, v0$\begin{array}{}{\stackrel{˙}{V}}_{+}^{r}\cap {L}^{\mathrm{\infty }}\left(\mathit{\Omega }\right)\end{array}$:

$∑i=1N∫Ω(−∂xi(bi(x,∂xiv))vr−1(x)+∂xi(bi(x,∂xiv0))v0r−1(x))(vr−v0r) dx≥0$

#### Proof

We apply Theorem 1.2. For A = Bi : Ω × ℝ → ℝ and by replacing ∇ by xi. □

## 3 Application of Picone identity to quasilinear elliptic equations

The aim of this section is to establish Theorem 1.3.

## 3.1 Preliminary results

The first lemma is the Picone identity in the context of the p(x)-Laplacian operator.

#### Lemma 3.1

Let r ∈ [1, p] and u, v$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$ (Ω) ∩ L(Ω) two positive functions. Then for any xΩ

$|∇u|p(x)+|∇v|p(x)≥|∇v|p(x)−2∇v.∇(urvr−1)+|∇u|p(x)−2∇u.∇(vrur−1).$

Following the proof of Theorem 1.1 in [19], we first prove the following comparison principle:

#### Lemma 3.2

Let λ ≥ 0 and u, v$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Lα(x)(Ω) two nonnegative functions for some function α ∈ 𝓟(Ω) satisfying 1 < αα+ < ∞. Assume for any ϕ$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω), ϕ ≥ 0:

$∫Ω|∇u|p(x)−2∇u.∇ϕ+uα(x)−1χu≥λϕ dx≥∫Ω|∇v|p(x)−2∇v.∇ϕ+vα(x)−1χv≥λϕ dx$

where

$χv≥λ(x)=1if λ≤v<∞;0if 0≤v<λ,$

and uv a.e. in ∂ Ω. Then uv a.e. in Ω.

#### Proof

Let ϕ = (vu)+$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) and Ω1 = {xΩ : u(x) < v(x)}. Then

$0≤−∫Ω1(|∇u|p(x)−2∇u−|∇v|p(x)−2∇v).∇(u−v) dx−∫Ω1(uα(x)−1χu≥λ−vα(x)−1χv≥λ)(u−v) dx≤0$

from which we obtain uv a.e. in Ω. □

Using lemma 3.2, we show the following strong maximum principle:

#### Lemma 3.3

Let h, lL(Ω) be nonnegative functions, h > 0 and k : Ω × ℝ+ → ℝ+. Let α, β ∈ 𝓟(Ω) be two functions such that 1 < ββ+ < αα+ < ∞. Let uC1(Ω) be nonnegative and a nontrivial solution to

$−Δp(x)u+l(x)uα(x)−1=h(x)uβ(x)−1+k(x,u)inΩ;u=0on∂Ω.$(3.1)

• (c1)

$\begin{array}{}\frac{l}{h}\end{array}$L(Ω)

or

• (c2)

k : Ω × ℝ+ → ℝ+ satisfying $\begin{array}{}\underset{t\to {0}^{+}}{lim inf}\frac{k\left(x,t\right)}{{t}^{\alpha \left(x\right)-1}}>\parallel l{\parallel }_{{L}^{\mathrm{\infty }}}\end{array}$ uniformly in x.

Then u is positive in Ω.

#### Proof

We follow the idea of the proof of Theorem 1.1 in [19]. For the reader’s convenience we have included the detailed proof. We rewrite our equation (3.1) under condition (c1) as follows:

$−Δp(x)u+l(x)uα(x)−1χu≥λ≥h(x)uβ(x)−1(1−χu≥λ)1−l(x)h(x)uα(x)−β(x),$

since $\begin{array}{}\frac{l}{h}\end{array}$L(Ω), we choose λ ∈ (0, 1) small enough such that for any u(x) ≤ λ, we have 1 − $\begin{array}{}\frac{l\left(x\right)}{h\left(x\right)}{u}^{\alpha \left(x\right)-\beta \left(x\right)}\ge 1-\parallel \frac{l}{h}{\parallel }_{{L}^{\mathrm{\infty }}\left(\mathit{\Omega }\right)}{\lambda }^{{\alpha }_{-}-{\beta }_{+}}\ge 0\end{array}$.

Assuming condition (c2), we have

$−Δp(x)u+l(x)uα(x)−1χu≥λ≥k(x,u)−(1−χu≥λ)l(x)uα(x)−1.$

We choose λ small enough such that for any u(x) ≤ λ, we have k(x, u) − l(x) uα(x)−1 ≥ 0. Hence under both conditions, we get for any xΩ,

$−Δp(x)u+l(x)uα(x)−1χu≥λ≥0.$

Suppose that there exists x1 such that u(x1) = 0 then using the fact that u is nontrivial, we can find x2Ω and a ball B(x2, 2C) in Ω such that x1 B(x2, 2C) and u > 0 in B(x2, 2C).

Let a = ∈ f{u(x) : |xx2| = C} then a > 0 and choosing x2 close enough to x1 such that 0 < a < λ and ∇ u(x1) = 0 since u(x1) = 0.

Denote the annulus P = {xΩ : C < |xx2| < 2C}. We define p1 = p(x1), M = sup{|∇ p(x)| : xP}, b = 8M + 2, l1 = −b ln $\begin{array}{}\left(\frac{a}{C}\right)+\frac{2\left(N-1\right)}{C}\end{array}$ and

$j(t)=ael1Cp1−1−1(el1tp1−1−1) ∀ t∈[0,C].$

We have

$aCe−l1Cp1−1

and then

$(aC)3≤j′(t)≤1 ∀ t∈[0,C].$(3.2)

We choose C < 1 and using ∇ u(x1) = 0, $\begin{array}{}\frac{a}{C}\end{array}$ < 1 small enough such that for any xP

$p(x)−1p1−1≥12.$(3.3)

Without loss of generality we can take x2 = 0 and we set r = |xx2| = |x|, t = 2Cr. For t ∈ [0, C] and r ∈ [C, 2C], denote w(r) = j(2Cr) = j(t), then

$w′(r)=−j′(t),w″(t)=j″(t).$

From (3.2) and (3.3), we obtain

$div(|∇w|p(x)−2∇w)=(p(x)−1))(j′(t))p(x)−2j″(t)−N−1r(j′(t))p(x)−1−(j′(t))p(x)−1ln⁡(j′(t))∑i=1n∂p∂xi.xir≥(j′(t))p(x)−1(12l1+Mln⁡(j′(t))−N−1r)≥−ln⁡(aC)(j′(t))p(x)−1≥0.$

Since j(t) < a < λ , we deduce

$−div(|∇w|p(x)−2∇w)+wα(x)−1χw≥λ≤0.$

On ∂ P, w(C) = j(C) = au(x) and w(2C) = j(0) = 0 ≤ u(x). Then by Lemma 3.2, we obtain wu on P. Finally,

$lims→0+u(x1+s(x2−x1))−u(x1)s≥lims→0+w(x1+s(x2−x1))−w(x1)s=j′(0)>0$

which contradicts ∇ u(x1) = 0. Therefore, u > 0 in Ω. □

#### Remark 3.1

Conditions (c1) and (c2) can be replaced by the condition that there exists t0 such that h(x)tβ(x)−1 + k(x, t) − l(x) tα(x)−1 ≥ 0 for all 0 < t < t0 and xΩ.

#### Lemma 3.4

Under the same conditions of h, l, k as in Lemma 3.3, let uC1(Ω) be the nonnegative and nontrivial solution of (3.1), x1∂ Ω , u(x1) = 0 and Ω satisfies the interior ball condition at x1, then $\begin{array}{}\frac{\mathrm{\partial }u}{\mathrm{\partial }\stackrel{\to }{n}}\left({x}_{1}\right)\end{array}$ < 0 where n⃗ is the outward unit normal vector at x1.

#### Proof

Choose C > 0 small enough such that B(x2, 2C) ⊂ Ω, x1 B(x2, 2C). Then x2 = x1 + 2 C n⃗, where n⃗ is the outward normal at x1. Denote P = {xΩ : C < |xx2| < 2C} and by choosing a such that 0 < a < λ, then by Lemma 3.3, there exist a subsolution wC1(P) ∩ C2(P) of (3.1) in P and w satisfies wu in P with w(x1) = 0, $\begin{array}{}\frac{\mathrm{\partial }w}{\mathrm{\partial }\stackrel{\to }{n}}\left({x}_{1}\right)\end{array}$ < 0. Hence, we get $\begin{array}{}\frac{\mathrm{\partial }u}{\mathrm{\partial }\stackrel{\to }{n}}\left({x}_{1}\right)\le \frac{\mathrm{\partial }w}{\mathrm{\partial }\stackrel{\to }{n}}\left({x}_{1}\right)<0\end{array}$. □

## 3.2 Proof of Theorem 1.3

#### Proof of Theorem 1.3

We perform the proof along five steps. First we introduce notations. Define F, G : Ω × ℝ → ℝ+ as follows:

$F(x,t)=h(x)q(x)tq(x) if 0≤t<∞;0 if −∞

and

$G(x,t)=l(x)s(x)ts(x) if 0≤t<∞;0 if −∞

We also extend the domain of f and g to all Ω × ℝ by setting

$f(x,t)=∂F∂t(x,t)=0 and g(x,t)=∂G∂t(x,t)=0 for (x,t)∈Ω×(−∞,0).$

Define the energy functional 𝓔 : $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω) → ℝ by

$E(u)=∫Ω|∇u|p(x)p(x) dx+∫ΩG(x,u(x)) dx−∫ΩF(x,u(x)) dx.$(3.4)

• Step 1

Existence of a global minimizer

Since $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ↪ Lq(x)(Ω) (see Theorem 3.3.1 and Theorem 8.2.4 in [9]), the functional 𝓔 is well-defined for every function u$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω).

For $\begin{array}{}\parallel u{\parallel }_{{W}_{0}^{1,p\left(x\right)}}\end{array}$ large enough: by (1.3) or (1.4)

$E(u)≥∫Ω|∇u|p(x)p(x)−∫Ωh(x)q(x)|u|q(x)≥1p−∥∇u∥Lp(x)p−−Cρq(u)≥1p−∥u∥W01,p(x)p−−C∥u∥W01,p(x)q~$

where $\begin{array}{}\stackrel{~}{q}=\left\{\begin{array}{cc}{q}_{-}& \text{\hspace{0.17em}if\hspace{0.17em}}\parallel u{\parallel }_{{L}^{p\left(x\right)}}\le 1\\ {q}_{+}& \text{\hspace{0.17em}if\hspace{0.17em}}\parallel u{\parallel }_{{L}^{p\left(x\right)}}>1\end{array}\right\\end{array}$ Since p > q+, this implies

$E(u)→∞ as ∥u∥W01,p(x)→+∞.$

We argue similarly when |u|Ls(x) → ∞ and we deduce 𝓔 is coercive. The continuity of 𝓔 on $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω) is given by Theorem 3.2.8 and 3.2.9 of [9]. Hence we get the existence of at least one global minimizer, say u0, to (3.4).

• Step 2

Claim: u0 ≥ 0 and u0 ≢ 0

Since u0 is a global minimizer of 𝓔 then 𝓔($\begin{array}{}{u}_{0}^{+}\end{array}$) ≥ 𝓔(u0) where $\begin{array}{}{u}_{0}^{+}\end{array}$ = max{u0, 0} ∈ $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω). Set Ω = {xΩ : u0(x) < 0}. We have

$E(u0)=∫Ω|∇u0|p(x)p(x) dx+∫ΩG(x,u0(x)) dx−∫ΩF(x,u0(x)) dx=E(u0+)+∫Ω−|∇u0|p(x)p(x) dx$

which implies $\begin{array}{}\underset{{\mathit{\Omega }}^{-}}{\int }\frac{|\mathrm{\nabla }{u}_{0}{|}^{p\left(x\right)}}{p\left(x\right)}=0\end{array}$ i.e. ∇u0(x) = 0 a.e. in Ω then by (1.3) and (1.4) we have u0 = 0 a.e in Ω. This implies that u0 ≥ 0.

In order to show that u0 ≢ 0 in Ω, we construct a function v in $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω)∩ L(Ω) such that 𝓔(v) < 0 = 𝓔(0). Precisely, consider v = where ϕ$\begin{array}{}{C}_{c}^{1}\end{array}$(Ω), ϕ ≥ 0, ϕ ≢ 0 in Ω and for 0 < t ≤ 1 small enough, we have

$E(v)≤tq+(c1tp−−q++c2ts−−q+−c3)$

where for any i ∈ {1, 2, 3}, ci are suitable constants independent of t. Hence, choosing t small enough the right-hand side is negative and we conclude that 𝓔() < 0 = 𝓔(0) which implies u0 ≢ 0.

• Step 3

u0 satisfies the equation in (1.7)

Since u0 is a global minimizer and 𝓔 is C1 on $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω), then for any ϕ$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω), we have

$〈E′(u0),ϕ〉=∫Ω|∇u0|p(x)−2∇u0.∇ϕ dx−∫Ωf(x,u0)ϕ dx+∫Ωg(x,u0)ϕ dx=0.$

• Step 4

Regularity and positivity of weak solutions

First we prove that all nonnegative weak solutions of (1.7) belongs to L(Ω) which yields C1, α(Ω) regularity.

Let K(x, t) = h(x) tq(x)–1l(x) ts(x)–1 and $\begin{array}{}\mathit{\Lambda }\text{\hspace{0.17em}}\stackrel{\text{def}}{=}\text{\hspace{0.17em}}max\left\{{\left|\left|\frac{h}{l}\right|\right|}_{{L}^{\mathrm{\infty }}},1{\right\}}^{1/\left({s}_{-}-{q}_{+}\right)}\end{array}$

Then it is not difficult to show that for any tΛ, K(x, t) ≤ 0.

Let u be a nonnegative function satisfying weakly the equation in (1.7). Then for any ϕ$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω),

$∫Ω|∇u|p(x)−2∇u.∇ϕ dx=∫Ω(h(x)uq(x)−1−l(x)us(x)−1)ϕ(x) dx.$

Taking the testing function ϕ(x) = (uΛ)+, we get

$∫Ω|∇(u−Λ)+|p(x)≤0.$

By using (1.4), we deduce $\begin{array}{}\parallel \left(u-\mathit{\Lambda }{\right)}^{+}{\parallel }_{{W}_{0}^{1,p\left(x\right)}}=0\end{array}$ which implies u(x) ≤ Λ.

From Theorem 1.2 in [10], we get uC1,α(Ω) for some α ∈ (0, 1).

Furthermore assuming $\begin{array}{}x\to \frac{l\left(x\right)}{h\left(x\right)}\end{array}$ belongs to L(Ω), Lemma 3.3 yields u > 0 in Ω.

• Step 5

Uniqueness of the positive solution of (1.7)

Let u, v be two positive solutions of (1.7). Thus for any ϕ, ϕ̃$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ Ls(x)(Ω),

$∫Ω|∇u|p(x)−2∇u.∇ϕ dx=∫Ω(h(x)uq(x)−1−l(x)us(x)−1)ϕ(x) dx$

and

$∫Ω|∇v|p(x)−2∇v.∇ϕ~ dx=∫Ω(h(x)vq(x)−1−l(x)vs(x)−1)ϕ~(x) dx.$

By the previous steps, u and v belong to C1(Ω) and Lemma 3.4 implies u, v$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$. Hence taking the testing functions as $\begin{array}{}\varphi =\frac{\left({u}^{{p}_{-}}-{v}^{{p}_{-}}{\right)}^{+}}{{u}^{{p}_{-}-1}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\stackrel{~}{\varphi }=\frac{\left({v}^{{p}_{-}}-{u}^{{p}_{-}}{\right)}^{-}}{{v}^{{p}_{-}-1}}\in {W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\end{array}$ (with the following notation t$\begin{array}{}\stackrel{\text{def}}{=}\end{array}$ max{0, –t}) and from Lemma 3.1 we obtain

$0≤∫{u>v}(|∇u|p(x)−2∇u−|∇v|p(x)−2∇v).∇(u−v)dx=∫{u>v}h(x)(uq(x)−p−−vq(x)−p−)(up−−vp−) dx+∫{u>v}l(x)(vs(x)−p−−us(x)−p−)(up−−vp−) dx.$

Since q+ps, the both terms in right-hand side are nonpositive. This implies v(x) ≥ u(x) a.e in Ω.

Finally reversing the role of u and v, we get u = v.□

#### Remark 3.2

Theorem 1.3 still holds when the condition $\begin{array}{}\frac{l}{h}\end{array}$L(Ω) is replaced by p+ < s and using strong maximum principle in [19].

## 4 Application to Fast diffusion equations

In this section, we establish Theorems 1.4 and 1.5. To this aim, we use a time semi-discretization method associated to (1.9). With the help of accurate energy estimates about the related quasilinear elliptic equation and passing to the limit as the discretization parameter goes to 0, we prove the existence and the properties of weak solutions to (1.8). In the subsection below, we study the associated elliptic problem.

## 4.1 Study of the quasilinear elliptic problem associated to F.D.E.

Consider the following problem

$v2q−1−λΔp(x)v=h0(x)vq−1+λf(x,v) in Ω;v>0 in Ω;v=0 on ∂Ω.$(4.1)

Assume h0L(Ω)+ and f satisfies (f1)-(f3). Then from (f3), we have (f0) : $\begin{array}{}\underset{s\to +\mathrm{\infty }}{lim}\frac{f\left(x,s\right)}{{s}^{{p}_{-}-1}}=0\end{array}$ uniformly in xΩ. Therefore, for any ϵ > 0, there exists a positive constant Cϵ such that for any (x, s) ∈ Ω × ℝ+:

$0≤f(x,s)≤Cϵ+ϵsp−−1.$(4.2)

We have the following preliminary result about (4.1):

#### Theorem 4.1

Let λ > 0, q ∈ (1, p], f : Ω × ℝ+ → ℝ+ satisfying (f0) and (f1) and h0L(Ω)+. Then there exists a weak solution vC1(Ω) to (4.1), i.e. for any ϕW $\begin{array}{}\stackrel{\text{def}}{=}\end{array}$ $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ L2q(Ω)

$∫Ωv2q−1ϕ dx+λ∫Ω|∇v|p(x)−2∇v.∇ϕ dx=∫Ωh0vq−1ϕ dx+λ∫Ωf(x,v)ϕ dx.$(4.3)

In addition, if (f2) and (f3) hold then v$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$. Moreover if v1, v2$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$ are two weak solutions to (4.1) corresponding to h0 = h1, h2L(Ω)+ respectively, then we have

$∥(v1q−v2q)+∥L2≤∥(h1−h2)+∥L2.$(4.4)

#### Remark 4.1

(4.4) implies the uniqueness of the weak solution to (4.1) in $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$.

#### Proof

We perform the proof into several steps.

• Step 1

Existence of a weak solution

Consider the energy functional 𝓙 defined on W equipped with $\begin{array}{}\parallel .{\parallel }_{\mathbf{W}}=\parallel .{\parallel }_{{W}_{0}^{1,p\left(x\right)}}+\parallel .{\parallel }_{{L}^{2q}}\end{array}$

$J(v)=12q∫Ωv2q dx+λ∫Ω|∇v|p(x)p(x) dx−1q∫Ωh0D(v) dx−λ∫ΩF(x,v) dx$(4.5)

where

$D(t)=tq if 0≤t<∞;0 if −∞

We also extend the domain of f to all of Ω × ℝ by setting $\begin{array}{}f\left(x,t\right)=\frac{\mathrm{\partial }F}{\mathrm{\partial }t}\left(x,t\right)=0\end{array}$ for (x, t) ∈ Ω × (–∞, 0). From (4.2), Hölder inequality (1.5) and since $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$Lp(Ω), we obtain

$J(v)≥12q∥v∥L2q2q+λ∥v∥W01,p(x)p−−1q∥h0∥L2∥v∥L2qq−λCϵ∫Ω|v|dx−λϵp−∫Ω|v|p− dx≥1q∥v∥L2qq(12∥v∥L2qq−∥h0∥L2)+λ∥v∥W01,p(x)((1−ϵ)∥v∥W01,p(x)p−−1−C~).$

Then by choosing ϵ small enough we conclude the coercivity of 𝓙 on W and 𝓙 is also continuous on W therefore we deduce the existence of a global minimizer v0 to 𝓙.

Furthermore we note

$J(v0)≥J(v0+)+12q∫Ω(v0−)2q dx+λ∫Ω|∇v0−|p(x)p(x) dx$

which implies v0 ≥ 0.

Now we claim that v0 ≢ 0 in Ω. Since 𝓙(0) = 0, it is sufficient to prove the existence of W such that 𝓙() < 0. For that take = where ϕ$\begin{array}{}{C}_{c}^{1}\end{array}$(Ω) is nonnegative function such that ϕ ≢ 0 and t > 0 small enough.

Since v0 is a global minimizer for the differentiable functional 𝓙, we have that v0 satisfies (4.3) i.e. v0 is a weak solution to (4.1). From Corollary A.1 we infer that v0L(Ω). Then by using Theorem A.2, we obtain, v0C1,α(Ω) for some α ∈ (0, 1).

From (f2) and Lemma 3.3 (with condition (c2)), we obtain v0 > 0 and by Lemma 3.4 we get $\begin{array}{}\frac{\mathrm{\partial }{v}_{0}}{\mathrm{\partial }\stackrel{\to }{n}}<0\end{array}$ on ∂Ω. Therefore, v0 belongs to $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$.

• Step 2

Contraction property (4.4)

Let v1 and v2 two positive weak solutions of (4.1) such that v1, v2$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$. For any ϕ, ΨW:

$∫Ωv12q−1ϕ dx+λ∫Ω|∇v1|p(x)−2∇v1.∇ϕ dx=∫Ωh1v1q−1ϕ dx+λ∫Ωf(x,v1)ϕ dx$

and

$∫Ωv22q−1Ψ dx+λ∫Ω|∇v2|p(x)−2∇v2.∇Ψ dx=∫Ωh2v2q−1Ψ dx+λ∫Ωf(x,v2)Ψ dx.$

Since $\begin{array}{}{v}_{1},{v}_{2}\in {C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+},\varphi ={\left({v}_{1}-\frac{{v}_{2}^{q}}{{v}_{1}^{q-1}}\right)}^{+}\text{and}\text{\hspace{0.17em}}\mathit{\Psi }={\left({v}_{2}-\frac{{v}_{1}^{q}}{{v}_{2}^{q-1}}\right)}^{-}\end{array}$ are well-defined and belong to W. Subtracting the two above expressions and using (f3) together with Lemma 3.1 we obtain

$∫Ω((v1q−v2q)+)2 dx≤∫Ω(h1−h2)(v1q−v2q)+ dx.$

Finally, applying the Hölder inequality we get (4.4).□

From Theorem 4.1, we deduce the accretivity of 𝓡:

#### Corollary 4.1

Let λ > 0, q ∈ (1, p], f : Ω × ℝ+ → ℝ+ satisfying (f1)-(f3) and h0L(Ω)+. Consider the following problem

$u+λRu=h0(x)inΩ;u>0inΩ;u=0on∂Ω.$(4.6)

Then there exists a unique distributional solution u ∈ 𝓓(𝓡) ∩ C1(Ω) of (4.6) i.e. ∀ϕ$\begin{array}{}{C}_{c}^{1}\end{array}$(Ω)

$∫Ωu0ϕ dx+λ∫Ω|∇u01/q|p(x)−2∇u01/q.∇(ϕu0(q−1)/q) dx=∫Ωh0ϕ dx+λ∫Ωf(x,u01/q)u0(q−1)/qϕ dx.$

Moreover, if u1 and u2 are two distributional solutions of (4.6) in 𝓓(𝓡) ∩ C1(Ω) associated to h1 and h2 respectively, then the operator 𝓡 satisfies

$∥(u1−u2)+∥L2≤∥(u1−u2+λ(Ru1−Ru2))+∥L2.$(4.7)

#### Proof

Define the energy functional 𝓔 on $\begin{array}{}{\stackrel{˙}{V}}_{+}^{q}\end{array}$L2(Ω) as 𝓔(u) = 𝓙(u1/q) where 𝓙 is defined in (4.5).

Let ϕ$\begin{array}{}{C}_{c}^{1}\end{array}$(Ω) and v0 is the global minimizer of (4.5) which is also the weak solution of (4.1) and u0 = $\begin{array}{}{v}_{0}^{q}\end{array}$ then there exists t0 = t0(ϕ) > 0 such that for t ∈ (–t0, t0), u0 + > 0. Hence we have

$0≤E(u0+tϕ)−E(u0)=12q(∫Ω(tϕ)2 dx+∫Ω2tu0ϕ dx)−1q∫Ωhtϕ dx+λ(∫Ω|∇(u0+tϕ)1/q|p(x)p(x) dx−∫Ω|∇u01/q|p(x)p(x) dx)−λ(∫ΩF(x,(u0+tϕ)1/q) dx−∫ΩF(x,u01/q) dx).$

Then divide by t and passing to the limits t → 0 we obtain u0 = $\begin{array}{}{v}_{0}^{q}\end{array}$ is the distributional solution of (4.6). Finally (4.7) and uniqueness follow from (4.4).□

We now generalize some above results for a larger class of potentials h0:

## 4.2 Further results for (4.1) and uniqueness

#### Theorem 4.2

Let λ > 0, f : Ω × ℝ+ → ℝ+ satisfying (f1)-(f3) and h0L2(Ω)+ and q ∈ (1, p]. Then there exists a positive weak solution vW of (4.1) in the sense of (4.3). Moreover assuming that h0 belongs to Lν(Ω) for some ν > $\begin{array}{}max\left\{1,\frac{N}{{p}_{-}}\right\},v\in {L}^{\mathrm{\infty }}\left(\mathit{\Omega }\right).\end{array}$

#### Proof

Let hn$\begin{array}{}{C}_{c}^{1}\end{array}$(Ω) such that hn ≥ 0 and hnh in L2(Ω). Define (vn) ⊂ C1,α(Ω) ∩ $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$ as for a fixed n, vn is the unique positive weak solution of (4.1) with h0 = hn i.e. vn satisfies: for ϕW

$∫Ωvn2q−1ϕ dx+λ∫Ω|∇vn|p(x)−2∇vn.∇ϕ dx=∫Ωhnvnq−1ϕ dx+λ∫Ωf(x,vn)ϕ dx.$(4.8)

Since (ab)2q ≤ (aqbq)2 for any q ≥ 1, (4.4) implies for any n, p ∈ ℕ*

$∥(vn−vp)+∥L2q≤∥(hn−hp)+∥L2q$

thus we deduce that (vn) converges to vL2q(Ω).

We infer that the limit v does not depend on the choice of the sequence (hn). Indeed, consider nhn such that nh0 in L2(Ω) and n the positive solution of (4.1) corresponding to n which converges to .

Then, for any n ∈ ℕ, (4.4) implies

$∥(vnq−v~nq)+∥L2≤∥(hn−h~n)+∥L2$

and passing to the limit we get v and then by reversing the role of v and we obtain v = .

So define, for any n ∈ ℕ*, hn = min{h, n}. Thus (vn) is nondecreasing and for any n ∈ ℕ*, vnv a.e. in Ω which implies vv1 > 0 in Ω.

We choose ϕ = vn in (4.8). Applying the Hölder inequality and (4.2), we obtain

$λ∫Ω|∇vn|p(x) dx≤∥hn∥L2∥vn∥L2qq+λCϵ∥vn∥L1+λϵ∥vn∥Lp−p−≤C+λϵ∥vn∥Lp−p−.$

Assume ∥∇vnLp(x) ≥ 1. Since $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ↪ Lp(Ω) and by (1.3) we deduce for some positive constant C > 0:

$λ∫Ω|∇vn|p(x) dx≤C+λϵC∫Ω|∇vn|p(x) dx.$

Choosing ϵ small enough and gathering with the case ∥∇vnLp(x) ≤ 1, we conclude (vn) is uniformly bounded in $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) and Lp(Ω). Hence vn converges weakly to v in $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) and by monotonicity of (vn) strongly in Lp(Ω) and in L2q(Ω). Taking now ϕ = vnv in (4.8), from (4.2) with ϵ = 1 and by Hölder inequality

$∫Ωf(x,vn)(vn−v) dx≤C∥vn−v∥L2q+∥vn∥Lp−p−−1∥vn−v∥Lp−→0$

and

$∫Ωhnvnq−1(vn−v) dx→0 and ∫Ωvn2q−1(vn−v) dx→0.$

Finally (4.8) becomes

$∫Ω|∇vn|p(x)−2∇vn.∇(vn−v) dx→0.$

Then, since vnv in $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω)

$∫Ω(|∇vn|p(x)−2∇vn−|∇v|p(x)−2∇v).∇(vn−v) dx→0.$

Lemma A.2 and Remark A.3 of [17] give the strong convergence of vn to v in $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω).

Since $\begin{array}{}\left({v}_{n}^{2q-1}\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\left({h}_{n}{v}_{n}^{q-1}\right)\end{array}$ are uniformly bounded in L2q/(2q–1)(Ω) and by (4.2), f(x, vn) is uniformly bounded in L2q/q–1(Ω) and f(x, vn) → f(x, v) a.e. in Ω. Then by Lebesgue dominated convergence theorem we have (up to a subsequence), for ϕW

$∫Ωvn2q−1ϕ dx→∫Ωv2q−1ϕ dx,∫Ωhnvnq−1ϕ dx→∫Ωhvq−1ϕ dx$

and

$∫Ωf(x,vn)ϕ dx→∫Ωf(x,v)ϕ dx.$

Finally we pass to the limit in (4.8) and we obtain v is a weak solution of (4.1). To conclude corollary A.1 implies vL(Ω).□

#### Remark 4.2

Let v1, v2 are the weak solutions of (4.1) obtained by Theorem 4.2 corresponding to h1, h2L2(Ω)+, h1h2 respectively. Then

$∥(v1q−v2q)+∥L2≤∥(h1−h2)+∥L2.$

#### Remark 4.3

As in Step 1 of the proof of Theorem 4.1, we can alternatively prove the existence of a weak solution by global minimization method.

Under the hypothesis of Theorem 4.2 and with the help of Picone identity, the following theorem gives the uniqueness of the solution to (4.1).

#### Theorem 4.3

Let v, be respectively a subsolution and supersolution to (4.1) for hLp0(Ω), p0 ≥ 2, h ≥ 0 and f satisfies (f1) and (f3). Then v.

#### Proof

We have for any nonnegative ϕ, ΨW

$∫Ωv2q−1ϕ dx+λ∫Ω|∇v|p(x)−2∇v.∇ϕ dx≤∫Ωhvq−1ϕ dx+λ∫Ωf(x,v)ϕ dx$

and

$∫Ωv~2q−1Ψ dx+λ∫Ω|∇v~|p(x)−2∇v~.∇Ψ dx≥∫Ωhv~q−1Ψ dx+λ∫Ωf(x,v~)Ψ dx.$

Subtracting the above inequalities with test functions $\begin{array}{}\varphi =\left(\frac{\left(v+ϵ{\right)}^{q}-\left(\stackrel{~}{v}+ϵ{\right)}^{q}}{\left(v+ϵ{\right)}^{q-1}}{\right)}^{+}\end{array}$ and $\begin{array}{}\mathit{\Psi }=\left(\frac{\left(\stackrel{~}{v}+ϵ{\right)}^{q}-\left(v+ϵ{\right)}^{q}}{\left(\stackrel{~}{v}+ϵ{\right)}^{q-1}}{\right)}^{-}\in \mathbf{W}\end{array}$ for ϵ ∈ (0, 1), we obtain

$∫{v>v~}(v2q−1(v+ϵ)q−1−v~2q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q) dx+λ∫{v>v~}|∇(v+ϵ)|p(x)−2∇(v+ϵ).∇((v+ϵ)q−(v~+ϵ)q(v+ϵ)q−1) dx+λ∫{v>v~}|∇(v~+ϵ)|p(x)−2∇(v~+ϵ).∇((v~+ϵ)q−(v+ϵ)q(v~+ϵ)q−1) dx≤∫{v>v~}h(vq−1(v+ϵ)q−1−v~q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q) dx+λ∫{v>v~}(f(x,v)(v+ϵ)q−1−f(x,v~)(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q) dx.$(4.9)

Since $\begin{array}{}\frac{\stackrel{~}{v}}{\stackrel{~}{v}+ϵ}\le \frac{v}{v+ϵ}<1\end{array}$ in {v > }, then we obtain

$(v2q−1(v+ϵ)q−1−v~2q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q)=(vq(vv+ϵ)q−1−v~q(v~v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q)≤vq((v+ϵ)q−(v~+ϵ)q)≤vq(v+ϵ)q≤vq(v+1)q.$

In the same fashion, we have

$0≤h(vq−1(v+ϵ)q−1−v~q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q)≤h(v+ϵ)q≤h(v+1)q.$

Moreover, as ϵ → 0

$(v2q−1(v+ϵ)q−1−v~2q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q)→(vq−v~q)2$

and

$h(vq−1(v+ϵ)q−1−v~q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q)→0$

a.e. in Ω. Then by Lebesgue dominated convergence theorem we have

$∫{v>v~}(v2q−1(v+ϵ)q−1−v~2q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q) dx→∫{v>v~}(vq−v~q)2 dx$(4.10)

and

$∫{v>v~}h(vq−1(v+ϵ)q−1−v~q−1(v~+ϵ)q−1)((v+ϵ)q−(v~+ϵ)q) dx→0.$(4.11)

Then by using Fatou’s Lemma and (f1), we have

$−lim infϵ→0∫{v>v~}f(x,v)(v+ϵ)q−1(v~+ϵ)q dx≤−∫{v>v~}f(x,v)vq−1v~q dx,−lim infϵ→0∫{v>v~}f(x,v~)(v~+ϵ)q−1(v+ϵ)q dx≤−∫{v>v~}f(x,v~)v~q−1vq dx$(4.12)

and

$∫{v>v~}f(x,v)(v+ϵ) dx→∫{v>v~}f(x,v)v dx,∫{v>v~}f(x,v~)(v~+ϵ) dx→∫{v>v~}f(x,v~)v~ dx.$(4.13)

By Lemma 3.1 we have,

$∫{v>v~}|∇(v+ϵ)|p(x)−2∇(v+ϵ).∇((v+ϵ)q−(v~+ϵ)q(v+ϵ)q−1) dx+∫{v>v~}|∇(v~+ϵ)|p(x)−2∇(v~+ϵ).∇((v~+ϵ)q−(v+ϵ)q(v~+ϵ)q−1) dx≥0.$(4.14)

Then plugging (4.10)-(4.14) and taking $\begin{array}{}\underset{ϵ\to 0}{lim sup}\end{array}$ in (4.9), we get by (f3)

$0≤∫{v>v~}(vq−v~q)2 dx≤λ∫{v>v~}(f(x,v)vq−1−f(x,v~)v~q−1)(vq−v~q) dx≤0.$

It implies v.□

#### Corollary 4.2

Let λ > 0, f : Ω × ℝ+ → ℝ+ satisfying (f1)-(f3) and h0L2(Ω)+Lγ(Ω) where γ > $\begin{array}{}max\left\{1,\frac{N}{{p}_{-}}\right\}.\end{array}$ Then there exists a unique positive distributional solution u ∈ 𝓓(𝓡) ∩ L(Ω) of (4.6) in the same sense as in Corollary 4.1.

Moreover if u1 and u2 are two positive distributional solutions of (4.6) for h1, h2L2(Ω)+ then 𝓡 satisfies

$∥(u1−u2)+∥L2≤∥(u1−u2+λ(Ru1−Ru2))+∥L2.$(4.15)

#### Proof

Define the functional energy 𝓔 on $\begin{array}{}{\stackrel{˙}{V}}_{+}^{q}\end{array}$L2(Ω) as 𝓔(u) = 𝓙(u1/q) where 𝓙 is given by (4.5).

By Theorem 4.2, Remark 4.3 and Theorem 4.3, v0 is the unique positive solution of (4.1) and then unique global minimizer of 𝓙. We take u0 = $\begin{array}{}{v}_{0}^{q}\end{array}$ and proceed as the proof of Corollary 4.1 and we obtain u0 = $\begin{array}{}{v}_{0}^{q}\end{array}$ is a distributional solution of (4.6). Finally Remark 4.2 gives (4.15).□

## 4.3 Existence of a weak solution to (1.8)

In this section, in light of Remark 1.3, we consider the problem (1.9) and establish the existence of weak solution when v0$\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω). Proof of Theorem 1.4: Let n* ∈ ℕ* and set Δt = T/n*. For 0 ≤ nn*, we define tn = t.

• Step 1

Approximation of h

For n ∈ {1, 2, … n*}, we define for t ∈ [tn–1, tn) and xΩ

$hΔt(t,x)=hn(x)=def⁡1Δt∫tn−1tnh(s,x)ds.$

Then by Jensen inequality,

$∥hΔt∥L2(QT)2=Δt∑n=1N∥hn∥L2(Ω)2=Δt∑n=1N1Δt∫tn−1tnh(s,x)dsL2(Ω)2≤∑n=1N∫tn−1tn∥h(s,.)∥L2(Ω)2ds≤∥h∥L2(QT)2.$

Hence hΔtL2(QT) and hnL2(Ω) and let ϵ > 0, then there exists a function hϵ$\begin{array}{}{C}_{0}^{1}\end{array}$(QT) such that $\begin{array}{}\parallel h-{h}_{ϵ}{\parallel }_{{L}^{2}\left({Q}_{T}\right)}<\frac{ϵ}{3}.\end{array}$

Hence,

$∥(hϵ)Δt−hΔt∥L2(QT)→0.$

Since ∥hϵ – (hϵ)ΔtL2(QT) → 0 as Δt → 0 then for small enough Δt we have

$∥hΔt−h∥L2(QT)≤∥(hϵ)Δt−hΔt∥L2(QT)+∥hϵ−(hϵ)Δt∥L2(QT)+∥h−hϵ∥L2(QT)<ϵ.$

Hence hΔth in L2(QT).

• Step 2

Time discretization of (1.9)

Define the following implicit Euler scheme and for n ≥ 1 , vn is the weak solution of

$(vnq−vn−1qΔt)vnq−1−Δp(x)vn=hnvnq−1+f(x,vn) in Ω;vn>0 in Ω;vn=0 on ∂Ω.$(4.16)

Note that the sequence (vn)n=1,2,…,n* is well-defined. Indeed for n = 1 the existence and the uniqueness of v1C1,α(Ω) ∩ $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$ follows from Theorems 4.1 and 4.3 with h = Δth1 + $\begin{array}{}{v}_{0}^{q}\end{array}$L(Ω)+. Hence by induction we obtain in the same way the existence and the uniqueness of the solution vn for any n = 2, 3, …, n* where vnC1,α(Ω) ∩ $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$.

• Step 3

Existence of a subsolution and supersolution

Now we construct a subsolution and a supersolution w and w of (4.16) such that for each n ∈ {0, 1, 2, …, n*}, vn satisfies 0 < wvnw.

Rewrite (4.16) as

$vn2q−1−ΔtΔp(x)vn=Δthn+vn−1qvnq−1+Δtf(x,vn).$(4.17)

Then following arguments in the proof of Theorems 4.1 and 4.3, from Theorem A.2 and from Lemma 3.4, for any μ > 0 there exists a unique weak solution, wμC1,α(Ω) ∩ $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$, to

$−Δp(x)w=μ(h0wq−1+f(x,w)) in Ω;w>0 in Ω;w=0 on ∂Ω.$(4.18)

Let μ1 < μ2 and wμ1, wμ2 be weak solutions of (4.18). Then,

$∫Ω|∇wμ1|p(x)−2∇wμ1.∇ϕ dx=μ1∫Ω(h0wμ1q−1+f(x,wμ1))ϕ dx,∫Ω|∇wμ2|p(x)−2∇wμ1.∇ψ dx=μ2∫Ω(h0wμ2q−1+f(x,wμ2))ψ dx.$

Subtracting the last two equations with $\begin{array}{}\varphi =\frac{\left({w}_{{\mu }_{1}}^{q}-{w}_{{\mu }_{2}}^{q}{\right)}^{+}}{{w}_{{\mu }_{1}}^{q-1}}and\psi =\frac{\left({w}_{{\mu }_{2}}^{q}-{w}_{{\mu }_{1}}^{q}{\right)}^{-}}{{w}_{{\mu }_{2}}^{q-1}}\in {W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\end{array}$ we obtain, by Lemma 3.1 and (f3), wμ1wμ2.

Then by using Theorems A.2 and A.3, we can choose μ small enough such that ∥wμC1,α(Ω)Cμ0 for all μμ0 and ∥wμL(Ω) → 0 as μ → 0. Therefore {wμ : μμ0} is uniformly bounded and equicontinuous in C1(Ω) and by Arzela Ascoli theorem ∥wμC1(Ω) → 0 as μ → 0 up to a subsequence. Then by mean value theorem we can choose μ small enough such that there exists wC1,α(Ω) ∩ $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$ such that 0 < w $\begin{array}{}\stackrel{\text{def}}{=}\end{array}$ wμv0. Also w is the subsolution of (4.17) for n = 1 i.e.

$∫Ωw_2q−1ϕ dx+Δt∫Ω|∇w_|p(x)−2∇w_.∇ϕ dx≤Δt∫Ω(h1w_q−1+f(x,w_))ϕ dx+∫Ωv0qw_q−1ϕ dx$

for all ϕ$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) and ϕ ≥ 0. We also recall v1 satisfies

$∫Ωv12q−1ψ dx+Δt∫Ω|∇v1|p(x)−2∇v1.∇ψ dx=Δt∫Ω(h1v1q−1+f(x,v1))ψ dx+∫Ωv0qv1q−1ψ dx$

for all ψ$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω).

By Theorem 4.3, we obtain, wv1 and then by induction a subsolution w such that 0 < wvn for all n = 0, 1, 2, …, n*.

Now we construct a supersolution. For that, we consider the following problem:

$−Δp(x)w=∥h∥L∞wq−1+f(x,w)+K in Ω;w>0 in Ω;w=0 on ∂Ω.$(4.19)

As above, there exists a unique weak solution to (4.19), wKC1(Ω) ∩ $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$. Let wK be the unique weak solution of

$−Δp(x)wK=K in Ω;wK=0 on ∂Ω.$(4.20)

From Theorem A.3, wKCK1/(p+–1+ν) dist(x, ∂Ω) where ν ∈ (0, 1) and ∥wKL(Ω) → ∞ as K → ∞. Then by weak comparison principle we can choose K large enough such that there exists such that v0wK < w$\begin{array}{}\stackrel{\text{def}}{=}\end{array}$ wK. We easily check that w is the supersolution of (4.17) for n = 1 i.e.

$∫Ωw¯2q−1ϕ dx+Δt∫Ω|∇w¯|p(x)−2∇w¯).∇ϕ dx≥Δt∫Ω(h1w¯q−1+K+f(x,w¯)ϕ dx+∫Ωv0qw¯q−1ϕ dx$

for all ϕ$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) and ϕ ≥ 0. From Theorem 4.3, we get wv1 and then by induction we have wvn for all n ∈ {1, 2, … n*}.

• Step 4

Energy estimates

Define the function for n = 1, … , n* and t ∈ [tn–1, tn)

$vΔt(t)=vn and v~Δt(t)=t−tn−1Δt(vnq−vn−1q)+vn−1q$

which satisfies

$vΔtq−1∂v~Δt∂t−Δp(x)vΔt=f(x,vΔt)+hnvΔtq−1.$(4.21)

Multiplying the equation (4.16) by $\begin{array}{}\frac{{v}_{n}^{q}-{v}_{n-1}^{q}}{{v}_{n}^{q-1}}\end{array}$ and summing from n = 1 to n′ ≤ n*, we get

$∑n=1n′∫ΩΔt(vnq−vn−1qΔt)2 dx+∑n=1n′∫Ω|∇vn|p(x)−2∇vn.∇(vnq−vn−1qvnq−1) dx=∑n=1n′∫Ωhn(vnq−vn−1q) dx+∑n=1n′∫Ωf(x,vn)vnq−1(vnq−vn−1q) dx.$

Then from Young inequality we have,

$∑n=1n′∫ΩΔt(vnq−vn−1qΔt)2 dx+∑n=1n′∫Ω|∇vn|p(x)−2∇vn.∇(vnq−vn−1qvnq−1) dx≤∑n=1n′Δt∥hn∥L22+14∑n=1n′∫ΩΔt(vnq−vn−1qΔt)2 dx+∑n=1n′Δtf(x,vn)vnq−1L22+14∑n=1n′∫ΩΔt(vnq−vn−1qΔt)2 dx$

i.e.

$12∑n=1n′∫ΩΔt(vnq−vn−1qΔt)2dx+∑n=1n′∫Ω|∇vn|p(x)−2∇vn.∇(vnq−vn−1qvnq−1) dx≤∑n=1n′Δt∥hn∥L22+∑n=1n′Δtf(x,vn)vnq−1L22.$

Using wvnw, from (4.2) and q$\begin{array}{}\frac{N}{2}\end{array}$ + 1, we obtain

$∫Ωf(x,vn)vn(q−1)2 dx≤C1∫Ω1dist2(q−1)(x,∂Ω)+dist2(p−−q)(x,∂Ω) dx≤C$(4.22)

where C is independent of n. Then by Step 1, we obtain

$(∂v~Δt∂t) is bounded in L2(QT) uniformly in Δt.$(4.23)

Now from Lemma 3.1, we have $|∇vn|p(x)−2∇vn.∇(vn−1qvnq−1)≤|∇vn−1|q|∇vn|p(x)−q≤qp(x)|∇vn−1|p(x)+(p(x)−q)p(x)|∇vn|p(x).$

Then we obtain for any n′ ≥ 1 $∑n=1n′Δt∥hn∥L22+∑n=1n′Δtf(x,vn)vnq−1L22≥∑n=1n′∫Ω|∇vn|p(x)−2∇vn.∇(vnq−vn−1qvnq−1) dx≥∑n=1n′∫Ω|∇vn|p(x) dx−∫Ωqp(x)|∇vn−1|p(x) dx−∫Ω(p(x)−q)p(x)|∇vn|p(x) dx≥q∫Ω|∇vn′|p(x)p(x) dx−q∫Ω|∇v0|p(x)p(x) dx$

which implies that $(vΔt) is bounded in L∞(0,T;W01,p(x)(Ω))uniformly in Δt.$(4.24)

Since $∇(v~Δt1/q)=1qζ∇vn(ζ+(1−ζ)(vn−1vn)q)(1−q)/q+(1−ζ)∇vn−1((1−ζ)+ζ(vnvn−1)q)(1−q)/q$

where $\begin{array}{}\zeta =\frac{t-{t}_{n-1}}{{\mathit{\Delta }}_{t}},\end{array}$ then we conclude that $(v~Δt1/q) is bounded in L∞(0,T;W01,p(x)(Ω)) uniformly in Δt.$(4.25)

Since $\begin{array}{}\left(\frac{{v}_{n}}{{v}_{n-1}}\right)\end{array}$ is uniformly bounded in L(Ω), $\begin{array}{}{v}_{{\mathit{\Delta }}_{t}}\stackrel{\ast }{⇀}v\text{\hspace{0.17em}and\hspace{0.17em}}{\stackrel{~}{v}}_{{\mathit{\Delta }}_{t}}^{1/q}\stackrel{\ast }{⇀}\stackrel{~}{v}\text{\hspace{0.17em}in\hspace{0.17em}}{L}^{\mathrm{\infty }}\left(0,T;{W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right)\right).\end{array}$ Furthermore using (4.23), we have $supt∈(0,T)∥v~Δt1/q−vΔt∥L2q(Ω)2q≤supt∈(0,T)∥v~Δt−vΔtq∥L2(Ω)2≤Δt→0 as Δt→0.$(4.26)

It follows from (4.26) that v = . By mean value theorem and (4.23), we get that (Δt)Δt is equicontinuous in C(0, T; Lr(Ω)) for 1 < r ≤ 2. Thus using wqΔtwq together with the interpolation inequality ∥.∥r$\begin{array}{}\parallel .{\parallel }_{\mathrm{\infty }}^{\alpha }\parallel .{\parallel }_{2}^{1-\alpha },\text{\hspace{0.17em}with\hspace{0.17em}}\frac{1}{r}=\frac{\alpha }{\mathrm{\infty }}+\frac{1-\alpha }{2},\end{array}$ we obtain that (Δt)Δt and $\begin{array}{}\left({\stackrel{~}{v}}_{{\mathit{\Delta }}_{t}}^{1/q}{\right)}_{{\mathit{\Delta }}_{t}}\end{array}$ is equicontinuous in C(0, T; Lr(Ω)) for any 1 < r < +∞. Again using interpolation inequality and Sobolev embedding, we get as Δt → 0+ and up to a subsequence that for all r > 1 $v~Δt→vq in C(0,T;Lr(Ω)),$(4.27)

and $vΔt→v in L∞(0,T;Lr(Ω)).$(4.28)

From (4.23) and (4.27), we obtain $∂v~Δt∂t→∂vq∂t in L2(QT).$(4.29)

• Step 5 :

v satisfies (1.10)

Multiplying (4.21) by (vΔtv) and integrating by parts, we get $∫0T∫ΩvΔtq−1∂v~Δt∂t(vΔt−v) dxdt+∫0T∫Ω|∇vΔt|p(x)−2∇vΔt.∇(vΔt−v) dxdt=∫0T∫Ωf(x,vΔt)(vΔt−v) dxdt+∫0T∫ΩhnvΔtq−1(vΔt−v) dxdt.$

From (4.28) and (4.29) , we have $∫0T∫ΩvΔtq−1∂v~Δt∂t(vΔt−v) dxdt+∫0T∫ΩhnvΔtq−1(vΔt−v) dxdt=oΔt(1)$

and from (4.24), (4.25), (4.28) and Lebesgue Dominated convergence theorem, $∫0T∫Ωf(x,vΔt)(vΔt−v) dx=oΔt(1).$

Then we obtain $∫0T∫Ω|∇vΔt|p(x)−2∇vΔt.∇(vΔt−v) dx→0 as Δt→0+.$

Then from [Step 4, Proof of Theorem 1.1, [12]] and from classical compactness argument we get $|∇vΔt|p(x)−2∇vΔt→|∇v|p(x)−2∇v in (Lp(x)/(p(x)−1)(QT))N.$(4.30)

From (4.26) and (4.27) we have, $∥vΔtq−1−vq−1∥L2(QT)≤∥vΔtq−1−vq−1∥L∞(0,T;L2)≤∥vΔtq−1−vq−1∥L∞(0,T;L2qq−1)≤∥vΔtq−vq∥L∞(0,T;L2)≤∥vΔtq−v~Δt∥L∞(0,T;L2)+∥v~Δt−vq∥L∞(0,T;L2)→0$(4.31)

as Δt → 0. By Hölder inequality we have for ϕ$\begin{array}{}{C}_{c}^{\mathrm{\infty }}\end{array}$(QT) $∫0T∫Ω(vΔtq−1∂v~Δt∂t−∂vq∂tvq−1)ϕ dx=∫0T∫ΩvΔtq−1(∂v~Δt∂t−∂vq∂t)ϕ dx+∫0T∫Ω∂vq∂t(vΔtq−1−vq−1)ϕ dx≤∥vΔtq−1ϕ∥L2(QT)∂v~Δt∂t−∂vq∂tL2(QT)+∥vΔtq−1−vq−1∥L2(QT)ϕ∂v~Δt∂tL2(QT)$

and $∫0T∫Ω(hnvΔtq−1−hvq−1)ϕ dx=∫0T∫Ωhn(vΔtq−1−vq−1)ϕ dx+∫0T∫Ω(hn−h)vq−1ϕ dx≤∥hnϕ∥L2(QT)∥vΔtq−1−vq−1∥L2(QT)+∥vq−1ϕ∥L2(QT)∥hn−h∥L2(QT).$

Then from (4.23), (4.28), (4.29), (4.31) and Step 1 we obtain $∫0T∫Ω(vΔtq−1∂v~Δt∂t−∂vq∂tvq−1)ϕ dx→0,∫0T∫Ω(hnvΔtq−1−hvq−1)ϕ dx→0 as Δt→0.$(4.32)

From (4.28) we have f(x, vΔt) → f(x, v) pointwise and from (4.24) together with (4.25) we have Ω f(x, vΔt) ϕ dx is bounded uniformly in Δt. Then by Lebesgue dominated convergence theorem we have $∫0T∫Ω(f(x,vΔt)−f(x,v))ϕ dx→0 as Δt→0.$(4.33)

Then finally gathering (4.30), (4.32) and (4.33), we conclude by passing to the limits in equation (4.21) that v is weak solution of (1.9). □

#### Remark 4.4

For q > $\begin{array}{}\frac{N}{2}\end{array}$+1, if f satisfies $\begin{array}{}\underset{s\to {0}^{+}}{lim}\frac{f\left(x,s\right)}{{s}^{\alpha }}=0\end{array}$ where α > q − 1 − $\begin{array}{}\frac{N}{2}\end{array}$ then Theorem 1.4 holds. Since wvnw then (4.22) is in this case $∫Ωf(x,vn)vnq−12 dx≤C1∫Ωw¯2αdist2(q−1)(x,∂Ω) dx+C2≤C$

where C is independent of n.

#### Remark 4.5

All the results in Section 4.1, Section 2 and Theorem 1.4 hold if we replace the assumption (f2) by hc > 0.

#### Proof of Theorem 1.5

For a given function g, let $\begin{array}{}\parallel g{\parallel }_{{2}^{+}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\stackrel{\text{def}}{=}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel \left[g{\right]}^{+}{\parallel }_{{L}^{2}\left(\mathit{\Omega }\right)}.\end{array}$ For z ∈ 𝓓(𝓡) and r, kL(QT)+ satisfying assumptions in Theorem 1.5, set $ϕ(t,s)=∥r(t)−k(s)∥2+∀(t,s)∈[0,T]×[0,T],$

for t ∈ [−T, T] $b(t,r,k)=∥u0q−z∥2++∥v0q−z∥2++|t|∥Rz∥2++∫0t+∥r(τ)∥2+dτ+∫0t−∥k(τ)∥2+dτ$

and $ψ(t,s)=b(t−s,r,k)+∫0sϕ(t−s+τ,τ)dτ if 0≤s≤t≤T∫0tϕ(τ,s−t+τ)dτ if 0≤t≤s≤T$

is a solution of $∂ψ∂t(t,s)+∂ψ∂s(t,s)=ϕ(t,s) on (t,s)∈[0,T]×[0,T];ψ(t,0)=b(t,r,k) on t∈[0,T];ψ(0,s)=b(−s,r,k) on s∈[0,T].$(4.34)

Define the following iterative scheme, u0 = $\begin{array}{}{u}_{0}^{q}\end{array}$ and for n ≥ 1 , un is the solution of $un−un−1Δt+Run=hn in Ω;un=0 on ∂Ω.$(4.35)

Note that the sequence {un}n = 1, 2, … , N is well defined. Indeed for n = 1 the existence and the uniqueness of u1 ∈ 𝓓(𝓡) follows from Corollary 4.1 with h = Δt h1 + u0L(Ω)+ and λ = Δt. Hence by induction we obtain in the same way the existence of the solution un for any n = 2, 3, … , N where un ∈ 𝓓(𝓡).

Moreover let denote by $\begin{array}{}\left({u}_{ϵ}^{n}\right)\end{array}$ the solution of (4.35) with Δt = ϵ , h = r, $\begin{array}{}{r}^{n}=\frac{1}{ϵ}\underset{\left(n-1\right)ϵ}{\overset{nϵ}{\int }}r\left(\tau ,.\right)d\tau \text{\hspace{0.17em}and\hspace{0.17em}}\left({u}_{\eta }^{m}\right)\end{array}$ the solution of (4.35) with Δt = η , h = k, $\begin{array}{}{k}^{m}=\frac{1}{\eta }{\int }_{\left(m-1\right)\eta }^{m\eta }k\left(\tau ,.\right)d\tau \end{array}$ respectively i.e we have $uϵn−uϵn−1ϵ+Ruϵn=rn;uηm−uηm−1η+Ruηm=km.$(4.36)

For (n, m) ∈ ℕ*, multiplying the equation in (4.36) by $\begin{array}{}\frac{ϵ\eta }{ϵ+\eta }\end{array}$ and then subtracting the two expressions we get, $ηη+ϵ(uϵn−uϵn−1)+ηϵη+ϵ(Ruϵn−Ruηm)−ϵη+ϵ(uηm−uηm−1)=ηϵη+ϵ(rn−km).$

Then we infer that $uϵn−uηm+ϵηϵ+η(Ruϵn−Ruηm)=ϵηϵ+η(rn−km)+ηϵ+η(uϵn−1−uηm)+ϵϵ+η(uϵn−uηm−1).$

Let $\begin{array}{}{\mathit{\Phi }}_{n,m}^{ϵ,\eta }=\parallel {u}_{ϵ}^{n}-{u}_{\eta }^{m}{\parallel }_{{2}^{+}}\end{array}$ and since 𝓡 satisfies (4.15) and setting $\begin{array}{}\lambda =\frac{ϵ\eta }{ϵ+\eta },\end{array}$ we get $Φn,mϵ,η=∥uϵn−uηm∥2+≤∥uϵn−uηm+ϵηϵ+η(Ruϵn−Ruηm)∥2+≤ϵηϵ+η∥rn−km∥2++ηϵ+η∥uϵn−1−uηm∥2++ϵϵ+η∥uϵn−uηm−1∥2+.$

Then by elementary calculations, we get $Φn,0ϵ,η=∥uϵn−uη∥2+≤b(tn,rϵ,kη)$

and $Φ0,mϵ,η≤b(−sm,rϵ,kη).$

Then by using above computations we get, $\begin{array}{}{\mathit{\Phi }}_{n,m}^{ϵ,\eta }\le {\psi }_{n,m}^{ϵ,\eta }\text{\hspace{0.17em}where\hspace{0.17em}}{\psi }_{n,m}^{ϵ,\eta }\end{array}$ satisfies $ψn,mϵη=ϵηϵ+η∥(rn−km)∥2++ηϵ+η∥ψn−1,mϵ,η∥2++ϵϵ+η∥ψn,m−1ϵ,η∥2+$

and $ψn,0ϵ,η=b(tn,rϵ,kη) and ψ0,mϵ,η=b(−sm,rϵ,kη).$

For (t, s) ∈ (tn−1, tn) × (sm−1, sm), set ϕϵ, η (t, s) = ∥rϵ(t) − kη(s)∥2+, $ψϵ,η=ψn,mϵ,η,bϵ,η(t,r,k)=b(tn,rϵ,kη),bϵ,η(−s,r,k)=b(−sm,rϵ,kη).$

Then by elementary calculations ψϵ, η satisfies the following discrete version of (4.34), $ψϵ,η(t,s)−ψϵ,η(t−ϵ,s)ϵ+ψϵ,η(t,s)−ψϵ,η(t,s−η)η=ϕϵ,η(t,s);ψϵ,η(t,0)=bϵ,η(t,r,k);ψϵ,η(0,s)=bϵ,η(s,r,k).$

Since rϵr in L2(QT) then bϵ, η (., r, k) → b(., r, k) in L([0, T]) and ϕϵ, ηϕ in L([0, T] ×[0, T]) and we deduce that ρϵ, η = ∥ψϵ, ηψL([0, T] ×[0, T]) → 0 (for more details see for instance [[20], Chapter 4, Lemma 4.3, page 136] and [[20], Chapter 4, Proof of Theorem 4.1, page 138]). Therefore, $∥uϵ(t)−uη(s)∥2+=Φϵ,η(t,s)≤ψϵ,η(t,s)≤ψ(t,s)+ρϵ,η.$

Since $\begin{array}{}{u}_{ϵ}\left(t\right)={v}_{ϵ}^{q}\left(t\right)\text{\hspace{0.17em}and\hspace{0.17em}}{u}_{\eta }\left(t\right)={v}_{\eta }^{q}\left(t\right),\end{array}$ we obtain $∥vϵq(t)−vηq(s)∥2+=Φϵ,η(t,s)≤ψϵ,η(t,s)≤ψ(t,s)+ρϵ,η.$(4.37)

From Theorem 1.4, $\begin{array}{}{v}_{ϵ}^{q}\text{\hspace{0.17em}and\hspace{0.17em}}{v}_{\eta }^{q}\end{array}$ satisfies 0 < w < vϵ, vη < w where w, w are subsolution and supersolution defined in (4.18) and (4.19) and $\begin{array}{}{v}_{ϵ}^{q}\to {v}_{1}^{q}\text{\hspace{0.17em}and\hspace{0.17em}}{v}_{\eta }^{q}\to {v}_{2}^{q}\end{array}$ a.e. in Ω where v1 and v2 are weak solutions of (1.9) with initial data u0, v0 respectively. Since $\begin{array}{}{v}_{ϵ}^{q}\to {v}_{1}^{q}\text{\hspace{0.17em}and\hspace{0.17em}}{v}_{\eta }^{q}\to {v}_{2}^{q}\end{array}$ in L(0, T;L2(Ω)) and passing to the limit in (4.37) as ϵ, η → 0 with t = s we get $∥v1q(t)−v2q(t)∥2+≤∥v1q(t)−vϵq(t)∥2++∥vηq(t)−v2q(t)∥2++∥vϵq(t)−vηq(t)∥2+≤∥u0q−z∥2++∥v0q−z∥2++∫0t∥r(γ)−k(γ)∥2+dγ.$

Then (1.11) follows since we can choose z arbitrary close to $\begin{array}{}{v}_{0}^{q}\end{array}$ and with r = h, k = g. □

## 5 An application to nonhomogeneous operators

In this final section, we prove Theorem 1.6. To this aim, we first study the properties of a related energy functional. Let m ≥ 1 and K : Ω × ℝN → ℝ+ be a continuous differentiable function which satisfies the following conditions:

• (k1)

KC1(Ω × ℝN) ∩ C2(Ω × ℝN ∖ {0}).

• (k2)

Ellipticity condition: ∃ k1 ≥ 0 and y ∈ (0, ∞) such that $∑i,j=1N∂2K∂ξi∂ξj(x,ξ)ηiηj≥y(k1+|ξ|)m−2|η|2.$

• (k3)

Growth condition: ∃ k2 ≥ 0 and Γ ∈ (0, ∞) such that $∑i,j=1N∂2K∂ξi∂ξj(x,ξ)≤Γ(k2+|ξ|)m−2$

for all ξ ∈ ℝN ∖ {0} and η ∈ ℝN.

#### Remark 5.1

From the assumption (k2), it follows that K is strictly convex and from (k1)-(k3) there exists some positive constant y1 and y2 with 0 < y1y2 < +∞ and some nonnegative constants Γ1 and Γ2 such that $y1|ξ|m−Γ1≤K(x,ξ)≤y2|ξ|m+Γ2$

for xΩ and ξ ∈ ℝN ∖ {0}.

Consider the associated functional 𝓙m defined by $Jm(u)=def∫Ω|u|p(x)p(x)K(x,∇uu)p(x)m dx.$

for any positive function $\begin{array}{}u\in {W}_{0}^{1,p\left(x\right)}\left(\mathit{\Omega }\right).\end{array}$ Now we extend Lemma 2.4 in [21] as follows:

#### Theorem 5.1

Let K : Ω × ℝN → ℝ+ satisfying (k1)-(k3) for some m ∈ [1, p]. Then, the function 𝓔 : $\begin{array}{}\stackrel{˙}{V}{}_{+}^{m}\end{array}$L(Ω) → ℝ+, defined by $\begin{array}{}\mathcal{E}\left(u\right)\phantom{\rule{thinmathspace}{0ex}}\stackrel{\text{def}}{=}\phantom{\rule{thinmathspace}{0ex}}{\mathcal{J}}_{m}\left({u}^{1/m}\right),\end{array}$ is ray-strictly convex (even strictly convex if p(⋅)≢ m).

#### Proof

We observe that for u$\begin{array}{}\stackrel{˙}{V}{}_{+}^{m}\end{array}$L(Ω) $E(u)=∫Ω1p(x)uKx,∇umup(x)m dx.$

Therefore, since for 1 ≤ mp, ttp(x)/m is convex in ℝ+ (even strictly convex if p(x) > m) it is enough to prove that $V˙+m∋u→uK(x,∇umu)$

is ray-strictly convex. To achieve this goal, let θ ∈ (0, 1) and u1, u2$\begin{array}{}\stackrel{˙}{V}{}_{+}^{m}\end{array}$ then by using the strict convexity of K we obtain, for xΩ $((1−θ)u1+θu2)K(x,(1−θ)∇u1+θ∇u2m((1−θ)u1+θu2))=((1−θ)u1+θu2)K(x,(1−θ)u1((1−θ)u1+θu2)∇u1mu1+θu2((1−θ)u1+θu2)∇u2mu2)≤((1−θ)u1+θu2)((1−θ)u1((1−θ)u1+θu2)K(x,∇u1mu1)+θu2((1−θ)u1+θu2)K(x,∇u2mu2))=(1−θ)u1K(x,∇u1mu1)+θu2K(x,∇u2mu2).$

The above inequality is always strict unless $\begin{array}{}\frac{\mathrm{\nabla }{u}_{1}}{{u}_{1}}=\frac{\mathrm{\nabla }{u}_{2}}{{u}_{2}},\end{array}$ i.e. u1/u2 ≡ Const. □

#### Proof of Theorem 1.6

Consider the functional 𝓙ϵ : $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$ (Ω) → ℝ , defined by $Jϵ(u)=∫Ω(|∇u|2+ϵu2)p(x)/2p(x) dx−∫ΩG(x,u) dx$

where the potential G(x, t) defined as $G(x,t)=∫0tg(x,s)ds if 0≤t<∞;0 if −∞

Assumptions (f1), () and Remark 5.1 ensure that 𝓙ϵ is well defined, coercive and continuous. Then there exists at least one global minimizer of 𝓙ϵ on $\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω), say u0. We can easily prove that u0 is nonnegative and nontrivial.

Since 𝓙ϵ is differentiable, we deduce that u0 is a weak solution of (1.12). Now from Theorems A.1 and A.2 in Appendix A, we obtain that any weak solution u to (1.12) belongs to C1, α(Ω) for some α ∈ (0, 1) and u > 0 in Ω and $\begin{array}{}\frac{\mathrm{\partial }u}{\mathrm{\partial }\stackrel{\to }{n}}\end{array}$ < 0 on ∂Ω. Therefore any weak solution belongs to $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$.

Now we prove that u0 is the unique weak solution to (1.12). Let W : $\begin{array}{}\stackrel{˙}{V}{}_{+}^{m}\end{array}$ → ℝ defined by $W(u)=Jϵ(u1/m)=∫Ω(|∇(u1/m)|2+ϵ(u1/m)2)p(x)/2p(x) dx−∫ΩG(x,u1/m) dx.$

The assumption () together with Theorem 5.1 with K(x, ξ) = (ϵ + |ξ|2)m/2 imply that W is strictly convex.

Let u1 a weak solution to (1.12). Then setting $\begin{array}{}{v}_{0}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\stackrel{\text{def}}{=}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{0}^{m},{v}_{1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\stackrel{\text{def}}{=}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{1}^{m}\in {\stackrel{˙}{V}}_{+}^{m}\end{array}$ and t ∈ [0, 1], we define $\begin{array}{}\xi \left(t\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\stackrel{\text{def}}{=}\end{array}$ Jϵ(((1 − t)v0+ t v1)1/m). Since u0 and u1 belong to $\begin{array}{}{C}_{d}^{0}\left(\overline{\mathit{\Omega }}{\right)}^{+}\end{array}$, ξ is differentiable in [0, 1]. From the convexity of 𝓔, we have for any t ∈ [0, 1] $ξ′(0)≤ξ′(t)≤ξ′(1).$(5.1)

Since u0 and u1 are weak solutions to (1.12), ξ′(0) = ξ′(1) = 0 and from (5.1) we get that ξ is constant which contradicts the strict convexity of 𝓔 unless u0u1. □

## A Appendix

In this section, we recall the following regularity of weak solutions of quasilinear elliptic differential equation $divA(x,u,Du)+B(x,u,Du)=0 on Ω;u=0 on ∂Ω.$(A.1)

Now we recall the boundedness and C0, α(Ω) regularity results of weak solutions of (A.1) satisfying the following growth conditions: $A(x,u,z)z≥a0|z|p(x)−b|u|r(x)−c;|A(x,u,z)|≤a1|z|p(x)−1+b|u|σ(x)+c;|B(x,u,z)|≤a2|z|α(x)+b|u|r(x)−1+c$(A.2)

where a0, a1, a2, b, c are positive constants and p* is the Sobolev embedding exponent of p and $p(x)≤r(x)(A.3)

#### Theorem A.1

([11], Theorem 4.1 and Theorem 4.4) Let (A.2)-(A.3) hold and p ∈ 𝓟log(Ω). If uW1, p(x)(Ω) is a weak solution of (A.1), then uC0, α({Ω).

Theorem A.2 below ensures C1, α(Ω) regularity to weak solutions of (A.1) under the additional assumptions on p, A and B:

Assumptions (Ak): A = (A1, A2, …, An) ∈ C(Ω × ℝ × ℝN, ℝN). For every (x, u) ∈ Ω × ℝ, A(x, u, .) ∈ C1 (ℝN ∖ {0}, ℝN), there exist a nonnegative constants k1, k2, k3 ≥ 0, a nonincreasing continuous function λ : [0, ∞) → (0, ∞) and a nondecreasing continuous function Λ : [0, ∞) → (0, ∞) such that for all x, x1, x2Ω, u, u1, u2 ∈ ℝ , η ∈ ℝN ∖ {0} and ξ = (ξ1, ξ2, …, ξn) ∈ ℝN, the following conditions are satisfied $A(x,u,0)=0,∑i,j∂Aj(x,u,η)∂ηi(x,u,η)ξiξj≥λ(|u|)(k1+|η|2)(p(x)−2)/2|ξ|2,∑i,j|∂Aj(x,u,η)∂ηi(x,u,η)|≤Λ(|u|)(k2+|η|2)(p(x)−2)/2 and |A(x1,u1,η)−A(x2,u2,η)|≤Λ(max{|u1|,|u2|})(|x1−x2|β1+|u1−u2|β2)×[(k+|η|2)(p(x1)−2)/2+(k+|η|2)(p(x2)−2)/2]|η|(1+|log⁡(k3+|η|2)|).$

Assumption (B): B :Ω × ℝ × ℝN → ℝ, the function B(x, u, η) is measurable in x and is continuous in (u , η), and $|B(x,u,η)|≤Λ(|u|)(1+|η|p(x)),∀(x,u,η)∈Ω¯×R×RN.$

#### Theorem A.2

(10], Theorem 1.2) Let assumptions (Ak), (B) hold. Assume p belongs to C0, β(Ω), for some β ∈ (0, 1). Suppose that Ω satisfy (Ω). If u$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ L(Ω) is a weak solution of (A.1), then uC1, α(Ω) where α ∈ (0, 1) anduC1, α(Ω) depends upon p, p+, λ(M), Λ(M), β1, β2, M, Ω where $\begin{array}{}M\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\stackrel{\text{def}}{=}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\parallel u{\parallel }_{{L}^{\mathrm{\infty }}\left(\mathit{\Omega }\right)}.\end{array}$

In the next theorem, we recall some results contained in Lemma 2.1 of [22] and Lemma 3.2 of [12]. Set ϱ = $\begin{array}{}\frac{{p}_{-}}{2|\mathit{\Omega }{|}^{1/N}{C}_{0}}\end{array}$ where C0 is the best embedding constant of $\begin{array}{}{W}_{0}^{1,1}\left(\mathit{\Omega }\right)\subset {L}^{\frac{N}{N-1}}\left(\mathit{\Omega }\right).\end{array}$

#### Theorem A.3

Let K > 0 and wK$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) ∩ C1, α(Ω) be the weak solution of (4.20).

Then for Kϱ, ∥wKL(Ω)C1 K1/(p−1), wK(x) ≥ C2 K1/(p+−1+ ς) dist(x, ∂Ω) where ς ∈ (0, 1) and for K < ϱ, ∥wKL(Ω)C3 K1/(p+−1) where C1, C2 and C3 depends upon p+, p, N, Ω. Moreover if K1 < K2 then wK1wK2.

Next we prove a slight extension of Proposition A.2 in [12].

#### Proposition A.1

Let pC(Ω) and q ∈ (1, p]. Assume uW satisfying for any ΨW: $∫Ω|∇u|p(x)−2∇u.∇Ψ dx=∫Ωhuq−1Ψ dx$(A.4)

where hL2(Ω) ∩ Lr(Ω) with $\begin{array}{}r>max\left\{1,\frac{N}{{p}_{-}}\right\}.\end{array}$ Then uL(Ω).

First we prove a regularity lemma.

#### Lemma A.1

Let u$\begin{array}{}{W}_{0}^{1,p}\end{array}$(Ω) satisfying for any BR, R < R0, and for all σ ∈ (0,1), and any kk0 > 0

$∫Ak,σR|∇u|pdx≤C∫Ak,Ru−kR(1−σ)p∗dx+kα|Ak,R|+|Ak,R|pp∗+ε+kβ|Ak,R|pp∗+ε+∫Ak,Ru−kR(1−σ)p∗dxpp∗|Ak,R|δ$

where Ak,R = {xBRΩu(x) > k}, 0 < α < $\begin{array}{}{p}^{\ast }=\frac{Np}{N-p},\end{array}$ β ∈ (1, p] and ε, δ > 0. Then uL(Ω).

#### Proof

A similar result exists in [23] or in [17] without the term $\begin{array}{}{k}^{\beta }|{A}_{k,R}{|}^{\frac{p}{{p}^{\ast }}+\epsilon }.\end{array}$ For the reader’s convenience, we include the complete proof.

Let x0Ω, BR the ball centred in x0. We define $\begin{array}{}{K}_{R}\stackrel{\text{def}}{=}{B}_{R}\cap \mathit{\Omega }\end{array}$ and we set

$rj=R2+R2j+1,r~j=rj+rj+12 and kj=k1−12j+1 for any j∈N.$

Define also

$Ij=∫Akj,rj|u(x)−kj|p∗dxandφ(t)=1if0≤t≤12,0ift≥34$

satisfying φC1([0, +∞)) and 0 ≤ φ ≤ 1. We set $\begin{array}{}{\phi }_{j}\left(x\right)=\phi \left(\frac{{2}^{j+1}}{R}\left(|x|-\frac{R}{2}\right)\right).\end{array}$ Hence φj = 1 on Brj+1 and φj = 0 on ℝNBj+1.

We have

$Ij+1=∫Akj+1,rj+1|u(x)−kj+1|p∗dx=∫Akj+1,rj+1|u(x)−kj+1|p∗φj(x)p∗dx≤∫KR(u(x)−kj+1)+φj(x))p∗dx.$

Since u$\begin{array}{}{W}_{0}^{1,p}\end{array}$(Ω), (ukj+1)+ φj$\begin{array}{}{W}_{0}^{1,p}\end{array}$(KR),

$Ij+1≲∫KR|∇((u−kj+1)+φj)|pdxp∗p≲∫Akj+1,r~j|∇u|pdx+∫Akj+1,r~j(u−kj+1)pdxp∗p$

where we use the notation fg in the sense there exists a constant c > 0 such that fcg. Since j < rj, we have

$Ij+1≲2jp∗∫Akj+1,rj|u−kj+1|p∗dx+kj+1α|Akj+1,rj|+|Akj+1,rj|pp∗+ε+kj+1β|Akj+1,rj|pp∗+ε+2jp∫Akj+1,rj|u−kj+1|p∗dxpp∗|Akj+1,rj|δ+∫Akj+1,rj|u−kj+1|p∗dxp∗p.$(A.5)

Moreover, for any j, kjkj+1, this implies

$Ij≥∫Akj+1,rj|u−kj|p∗dx≥∫Akj+1,rj|kj−kj+1|p∗dx=|Akj+1,rj∥kj+1−kj|p∗.$

Then, for any k > k0 and j ∈ ℕ

$|Akj+1,rj|+kj+1p∗|Akj+1,rj|≲2jp∗Ij$

where the constant in the notation depends only on k0, p and α. From the previous inequality, we deduce

$kj+1β|Akj+1,rj|pp∗+ε≤kj+1p+εp∗|Akj+1,rj|pp∗+ε≲2j(p+εp∗)Ijpp∗+ε.$

Replacing in (A.5), we obtain

$Ij+1≲(2jp∗Ij+2j(p+εp∗)Ijpp∗+ε+2j(p+δp∗)Ijpp∗+δ)p∗p.$(A.6)

Setting $\begin{array}{}M=\frac{p}{{p}^{\ast }}max\left\{{p}^{\ast },\phantom{\rule{thinmathspace}{0ex}}p+\epsilon {p}^{\ast },\phantom{\rule{thinmathspace}{0ex}}p+\delta {p}^{\ast }\right\}\end{array}$ and $\begin{array}{}\theta =min\left\{1-\frac{p}{{p}^{\ast }},\phantom{\rule{thinmathspace}{0ex}}\epsilon ,\phantom{\rule{thinmathspace}{0ex}}\delta \right\}\end{array}$ and noting

$Ij≤∫KR(|u−kj|+)p∗dx≤∫KR|u|p∗≤∥u∥W01,pp∗,$

(A.6) becomes

$Ij+1≲2jMIj1+θp∗p$

where the constant depends on $\begin{array}{}\parallel u{\parallel }_{{W}_{0}^{1,p}},\phantom{\rule{thinmathspace}{0ex}}{k}_{0},\phantom{\rule{thinmathspace}{0ex}}\alpha \end{array}$ and p. We conclude with Lemma 4.7 in Chapter 2 of [24].

For this it suffices to prove that I0 is small enough. Indeed uLp*(Ω) implies

$I0=∫Ak2,R|u−k2|p∗dx→0 as k→∞.$

Hence for k large enough, $\begin{array}{}{I}_{0}\le {C}^{-\frac{1}{\eta }}\left({2}^{M}{\right)}^{-\frac{1}{{\eta }^{2}}}\end{array}$ with $\begin{array}{}\eta =\frac{\theta {p}^{\ast }}{p}.\end{array}$ Thus Ij converges to 0 as j → +∞ and

$∫Ak,R2|u−k|p∗dx=0.$

We deduce that uk on $\begin{array}{}{K}_{\frac{R}{2}}.\end{array}$ In the same way, we prove that −uk on $\begin{array}{}{K}_{\frac{R}{2}}.\end{array}$

Since Ω is compact, we conclude that uL(Ω).□

#### Proof of Proposition A.1:

We follow the idea of the proof of Theorem 4.1 in [11].

Let x0Ω, BR the ball of radius R centered in x0 and $\begin{array}{}{K}_{R}\stackrel{\text{def}}{=}\mathit{\Omega }\cap {B}_{R}.\end{array}$ We define

$p+=defmaxKRp(x)andp−=defminKRp(x)$

and we choose R small enough such that p+ < (p)* where

$(p−)∗=defNp−N−p−if p−

Fix (s, t) ∈ $\begin{array}{}\left({\mathbb{R}}_{+}^{\ast }{\right)}^{2},\end{array}$ t < s < R then KtKsKR. Define φC(Ω), 0 ≤ φ ≤ 1 such that

$φ=1in Bt,0in RN∖Bs$

satisfying ∣∇φ∣ ≲ 1/(st). Let k ≥ 1, using the same notations as previously Ak,λ = {yKλu(y) > k} and taking Ψ = φp+(uk)+$\begin{array}{}{W}_{0}^{1,p\left(x\right)}\end{array}$(Ω) in (A.4), we obtain

$∫Ak,s|∇u|p(x)φp+dx+p+∫Ak,s|∇u|p(x)−2∇u⋅∇φφp+−1(u−k)+dx=∫Ak,shuq−1φp+(u−k)dx.$(A.7)

Hence by Young inequality, for ϵ > 0, we have

$p+∫Ak,s|∇u|p(x)−2∇u⋅∇φφp+−1(u−k)dx≤ε∫Ak,s|∇u|p(x)φ(p+−1)p(x)p(x)−1dx+cε−1∫Ak,s(u−k)p(x)|∇φ|p(x)dx.$

Since ∣∇φ∣ ≤ c/(st) and for any xKR, p+$\begin{array}{}\left({p}^{+}-1\right)\frac{p\left(x\right)}{p\left(x\right)-1},\end{array}$ we have $\begin{array}{}{\phi }^{\left({p}^{+}-1\right)\frac{p\left(x\right)}{p\left(x\right)-1}}\le {\phi }^{{p}^{+}}.\end{array}$ This implies

$p+∫Ak,s|∇u|p(x)−2∇u.∇φφp+−1(u−k)dx≤ε∫Ak,s|∇u|p(x)φp+dx+cε−1∫Ak,su−ks−tp(x)dx.$(A.8)

Using Hölder inequality, we estimate the right-hand side of (A.7) as follows:

$∫Ak,shuq−1φp+(u−k)dx≤∥h∥Lr∫Ak,sur(q−1)r−1(u−k)rr−1dxr−1r.$

Since $\begin{array}{}r>\frac{N}{{p}^{-}},\end{array}$ we have $\begin{array}{}\frac{\left({p}^{-}{\right)}^{\ast }}{{p}^{-}}\frac{r-1}{r}>1,\end{array}$ applying once again the Hölder inequality and the Young inequality, we obtain

$∫Ak,shuq−1φp+(u−k)dx≲∫Ak,suq(p−)∗p−dx+∫Ak,s(u−k)q(p−)∗p−dxp−(p−)∗|Ak,s|δ$

where $\begin{array}{}\delta =\frac{r-1}{r}-\frac{{p}^{-}}{\left({p}^{-}{\right)}^{\ast }}>0.\end{array}$

Set Ak,s,t = {xAk,su(x) − k > st} and its complement as $\begin{array}{}{A}_{k,s,t}^{c}.\end{array}$ Now we split the integrals in the right-hand side of (A.9) as follows:

$∫Ak,s,tu−ks−tq(p−)∗p−(s−t)q(p−)∗p−dx+∫Ak,s,tcu−ks−tq(p−)∗p−(s−t)q(p−)∗p−dx≲∫Ak,su−ks−t(p−)∗dx+|Ak,s|=defI$(A.9)

since q < p and we also have

$∫Ak,suq(p−)∗p−dx≲∫Ak,s(u−k)q(p−)∗p−+kq(p−)∗p−dx≲I+kq(p−)∗p−|Ak,s|.$

In the same way, the second term in the right-hand side of (A.8) can be estimated as follows:

$∫Ak,s∩Ak,s,tu−ks−tp(x)dx+∫Ak,s∩Ak,s,tcu−ks−tp(x)dx≲I.$(A.10)

Finally plugging (A.8)-(A.10), we obtain for ε small enough

$∫Ak,s|∇u|p(x)φp+dx≲I+|Ak,s|δ(I+kq(p−)∗p−|Ak,s|)p−(p−)∗$

where the constant depends on p, R and ε. Moreover we have

$(I+kq(p−)∗p−|Ak,s|)p−(p−)∗≲∫Ak,su−ks−t(p−)∗dxp−(p−)∗+|Ak,s|p−(p−)∗+kq|Ak,s|p−(p−)∗.$

To conclude, using the Young inequality, we obtain the following estimate:

$∫Ak,t|∇u|p−dx≤∫Ak,s|∇u|p(x)φp+dx≲∫Ak,su−ks−t(p−)∗dx+2|Ak,s|+(1+kq)|Ak,s|p−(p−)∗+δ+|Ak,s|δ∫Ak,su−ks−t(p−)∗dxp−(p−)∗.$

By Lemma A.1, we deduce that u bounded in Ω.□

Combining Theorem 4.1 of [11] and Proposition A.1, we have the following corollary:

#### Corollary A.1

Let pC(Ω̄) and q ∈ (1, p]. Assume uW and nonnegative satisfying for any ΨW, Ψ ≥ 0,

$∫Ωu2q−1Ψdx+∫Ω|∇u|p(x)−2∇u⋅∇Ψdx≤∫Ω(f(x,u)+huq−1)Ψdx$

where f verifies for any (x, t) ∈ Ω × ℝ+, ∣f(x, t)∣ ≤ c1 + c2ts(x)−1 with sC(Ω) such that for any xΩ, 1 < s(x) < p*(x) and hL2(Ω) ∩ Lr(Ω) with $\begin{array}{}r>max\left\{1,\frac{N}{{p}_{-}}\right\}.\end{array}$ Then uL(Ω).

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Accepted: 2018-11-29

Published Online: 2019-05-16

Published in Print: 2019-03-01

Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 327–360, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496,

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